Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth 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  < p ≤  and  < 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".

 #pragma comment(linker, "/STACK:1024000000,1024000000")

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<cmath>

 #include<math.h>

 #include<algorithm>

 #include<queue>

 #include<set>

 #include<bitset>

 #include<map>

 #include<vector>

 #include<stdlib.h>

 #include <stack>

 using namespace std;

 #define PI acos(-1.0)

 #define max(a,b) (a) > (b) ? (a) : (b)

 #define min(a,b) (a) < (b) ? (a) : (b)

 #define ll long long

 #define eps 1e-10

 #define N 1000000

 #define inf 1e12

 ll pow_mod(ll a,ll n,ll MOD)

 {

     if(n==)

        return %MOD;

     ll tt=pow_mod(a,n>>,MOD);

     ll ans=tt*tt%MOD;

     if(n&)

       ans=ans*a%MOD;

     return ans;

 }

 int main()

 {

    ll p,a;

    while(scanf("%I64d%I64d",&p,&a)==){

       if(p== && a==){

          break;

       }

       int flag=;

       for(int i=;i<(int)sqrt(p+0.5);i++){

          if(p%i==){

             flag=;

             break;

          }

       }

       if(flag==){

          printf("no\n");

          continue;

       }

       ll ans=pow_mod(a,p,p);

       //printf("%I64d\n",ans);

       if(ans==a){

          printf("yes\n");

       }else{

          printf("no\n");

       }

    }

     return ;

 }