A sanity check of the Digitally Encrypted challenges series.

```
____
Plain4 [o]-------\\ \
||xor|------(o) Ciph4
Key|>--//___/
____
Plain3 [o]-------\\ \
||xor|------(o) Ciph3
Key|>--//___/
____
Plain2 [o]-------\\ \
||xor|------(o) Ciph2
Key|>--//___/
____
Plain1 [o]-------\\ \
||xor|------(o) Ciph1
Key|>--//___/
____
Plain0 [o]-------\\ \
||xor|------(o) Ciph0
Key|>--//___/

0xD4C70F8A67D5456D---<|Key
```

Ciph[x] = Plain[x] xor 0xD4C70F8A67D5456D

Solving Script:

```python
key = 0xD4C70F8A67D5456D
c = [0xB6A46EE913B33E19, 0xBCA67BD510B43632, 0xA4B56AFE13AC1A1E, 0xBDAA7FE602E4775E, 0xEDF63AB850E67010]
print(b"".join(bytes.fromhex(hex(key ^ i)[2:]) for i in c))
```

Output:

```
b'bcactf{that_was_pretty_simple1239152735}'
```