Description

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

Input

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

Output

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

Sample Input

3 2

10 3

341 2

341 3

1105 2

1105 3

0 0

Sample Output

no

no

yes

no

yes

yes

---

题意：不是素数的数p，且a^p对p取模等于a，输出yes，其他的输出no。

直接暴力，不要打表，打表数组开不到1e9，暴力不会超时

#include
    
     #include
     
      #define ll long longll Pow(ll a,ll b){    ll res=1;    ll c=b;    while(b>0)    {        if(b&1) res=res*a%c;        a=a*a%c;        b>>=1;    }    return res;}ll su(ll n){    for(ll i=2;i*i<=n;i++)      {        if(n%i==0)        return 0;    }    return 1;  }int main(){    ll p,a,flag=0;    while(~scanf("%lld%lld",&p,&a))    {        if(a==0&&p==0) break;        if(su(p)==0) flag+=1;        if(flag!=0&&Pow(a,p)==a) printf("yes\n");        else printf("no\n");        flag=0;    }    return 0;}