چالش Sergio¶

pad = lambda s: '0'+s if len(s)%2  else s
long_to_bytes = lambda c: bytes.fromhex(pad(hex(c)[2:]))
inverse = lambda a, b: pow(a, -1, b)

e = 65537
N = 139551725550533062709001988886045836849
c = 70886407371304490355797613974913658205

p, q = 10073925425153076761, 13852765397894786009
assert p*q == N
phi = (p-1)*(q-1)
d = inverse(e, phi)
long_to_bytes(pow(c, d, N))