To solve this challenge we used Chinese remainder algorithm
   1) p and q are the primes   2) dp = d mod (p - 1)   3) dq = d mod (q - 1)   4) Qinv = 1/q mod p *This is not integer devision but multiplicative inverse   5) m1 = pow(c, dp, p)   6) m2 = pow(c, dq, q) 7-1) h = Qinv(m1 - m2) mod p ; if m1 < m2 7-2) h = Qinv * (m1 + q/p)   7) m = m2 + hq
flag{Theres_more_than_one_way_to_RSA}