PicoCTF crypto challenge - Mind your Ps and Qs

Description

In RSA, a small e value can be problematic, but what about N? Can you decrypt this? values

format :

C = ciphertext
p and q = prime numbers
n = p * q
phi = (p-1) * (q-1)
e = some number that 1 < e < phi and gcd(e,phi) == 1
d = e^(-1) mod phi

#! /usr/bin/env python3
from factordb.factordb import FactorDB
import gmpy2

c = 240986837130071017759137533082982207147971245672412893755780400885108149006
n = 83141682808041786634050496818899003281031619353365351602217578439972014107627
e = 65537

f = FactorDB(n)
f.connect()
p, q = f.get_factor_list()
ph = (p-1)*(q-1)
d = gmpy2.invert(e, ph)
plaintext = pow(c, d, n)
print("Flag: {}".format(bytearray.fromhex(format(plaintext, 'x')).decode()))