This is my code for RSA angstromCTF 2020.

rsa.py

#!/usr/bin/env python
def modinv(a, m) :
    a = a % m;
    for x in range(1, m) :
        if ((a * x) % m == 1) :
            return x
    return 1
    
n = 126390312099294739294606157407778835887
e = 65537
c = 13612260682947644362892911986815626931

p = 13536574980062068373
q = 9336949138571181619
phi = (p-1) * (q-1)

d = modinv(e, phi)
m = hex(pow(c,d,n))[2:-1]
print str(m).decode('hex')