> Encoding is not encryption, but what if I just encode the flag with base16,32,64? If I encode my precious flag for 150 times, surely no one will be able to decode it, right?

We have a file `onionlayerencoding.txt`, which is encoded in base16,32,64 150 times. We don't know the order, so we guess which decoding algorithm to use based on the characters of the string to the decode.

```
#!/usr/bin/env python3

from base64 import b64decode, b32decode, b16decode
from base64 import b64encode, b32encode, b16encode
import string
import sys

b64table = list(string.digits + string.ascii_uppercase + string.ascii_lowercase + '+/=')
b32table = list(string.ascii_uppercase + '234567=')
b16table = list(string.hexdigits.upper())

def is_b16(s):
return set(list(s)).issubset(set(b16table))

def is_b32(s):
return set(list(s)).issubset(set(b32table))

def is_b64(s):
return set(list(s)).issubset(set(b64table))

def main():
f = open("./onionlayerencoding.txt").read().encode()
for i in range(150):
try:
if is_b16(f.decode()):
sys.stdout.write('\rb16, i : ' + str(i))
f = b16decode(f)
continue
except:
pass
try:
if is_b32(f.decode()):
sys.stdout.write('\rb32, i : ' + str(i))
f = b32decode(f)
continue
except:
pass
try:
if is_b64(f.decode()):
sys.stdout.write('\rb64, i : ' + str(i))
f = b64decode(f)
continue
except:
pass
print("\n" + f.decode())

if __name__ == '__main__':
main()
```

Output:

`RITSEC{0n1On_L4y3R}`