[write-up][CTF(x)] [Crypto50] λ (lambda)

I used this program to encrypt a flag. The output was:n1s4_t1An(f1ctdb@mpl_h3)m3lp3y__Eas

print (lambda j,m:(lambda f,t:t if len(t) <= 1 else j([f(f,x)for x in m(j,m(reversed,(lambda s:zip(*[iter(s)]*(len(s)/2)))(t+"\x01"*(len(t)%2))))]))(lambda f,t:t if len(t) <= 1 else j([f(f,x)for x in m(j,m(reversed,(lambda s: zip(*[iter(s)]*(len(s)/2)))(t+"\x01"*(len(t)%2))))]),raw_input("Plaintext:")))(''.join,map).replace("\x01","")

I realized it's a obfuscation code so it too hard to reverse code to readable. So i think a ez way to solve this challenge. I think this program will change pos of char so i will find the pos of each char before encypt.
First i check length of cipher:
len("n1s4_t1An(f1ctdb@mpl_h3)m3lp3y__Eas")=35

Then i encrypt this type of flag : 


ctf(ABCDEFGHIJKLMNOPQRSTUVWXYZ0123)

after i encrypted this type of flag. i would found the pos of each char in this type of flag in encrypted flag. Here is my decrypt code :

def enc(plaintext): return (lambda j,m:(lambda f,t:t if len(t) <= 1 else j([f(f,x)for x in m(j,m(reversed,(lambda s:zip(*[iter(s)]*(len(s)/2)))(t+"\x01"*(len(t)%2))))]))(lambda f,t:t if len(t) <= 1 else j([f(f,x)for x in m(j,m(reversed,(lambda s: zip(*[iter(s)]*(len(s)/2)))(t+"\x01"*(len(t)%2))))]),plaintext))(''.join,map).replace("\x01","") def dcode(cipher): flag_type="ctf(ABCDEFGHIJKLMNOPQRSTUVWXYZ0123)" flag="" for x in flag_type: flag+=cipher[enc(flag_type).find(x)] return flag cip='n1s4_t1An(f1ctdb@mpl_h3)m3lp3y__Eas' print "\nFLAG : ", dcode(cip)

Input : n1s4_t1An(f1ctdb@mpl_h3)m3lp3y__Eas
FLAG : ctf(1@mbd4_1nsAn1ty_pl3asE_h3lp_m3)