Fermat's theorem states that for any prime number p and for any integer a > 1, a^p = a (mod p). That is, if we raise a to the p^th power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)

Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.

For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

#define _CRT_SECURE_NO_DEPRECATE

#include<iostream>

#include<algorithm>

#include<vector>

#include<cstring>

#include<string>

#include<bitset>

using namespace std;

#define INF 0x3f3f3f3f

#define MOD 1000000000

typedef long long ll;

const int N_MAX = ;

ll p, a;

ll ll_mult(ll a,ll x,ll p) {

    ll res = ;

        bitset<>tmp = static_cast<bitset<>>(x);//前面低位

        for (int i = ; i < tmp.size();i++) {

            if (tmp[i])res += a*( << i);

            res %= p;

        }

    return res;

}

ll mod_pow(ll x,ll n,ll p) {

    ll res = ;

    while (n) {

        if (n & )res = ll_mult(res,x,p);

        x = ll_mult(x, x,p);

        n >>= ;

    }

    return res;

}

bool is_prime(ll n) {

    for (int i = ; i*i <= n;i++) {

        if (n%i == )return false;

    }

    return n!=;

}

int main() {

    while (scanf("%lld%lld",&p,&a)&&(p||a)) {

        if (is_prime(p)) { puts("no"); continue; }

        if (mod_pow(a, p, p) == a)puts("yes");

        else puts("no");

    }

     return ;

}