Primitive Root 原根

题意：给定模P，n个c,判断c是否是p的原根，

《数论概论》中“幂模p与原根”一章中有提到阶的概念： 如果gcd(c,p)=1，则a模p的阶是指使得

a^d=1（mod p）的最小指数d(d>=1)； 例如2、3、4、5、6模7的阶分别是3、6、3、6、2。

重要性质：一个数a模p的阶e总能整除p-1。  所以可以枚举p-1的所有因子factor 

（不包括p-1），如果存

在小于p-1的因子满足a^factor=1(mod p) 则说明a模p的阶不是p-1；

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <queue>#include <vector>#include <stack>#include <algorithm>using namespace std;int a[10000];int mod,cnt;void pr(int p){     a[0]=1;cnt=0;     for(int i=2;i<=(int)sqrt(p);i++)     {           if(p%i==0)           {               a[++cnt]=i;               a[++cnt]=p/i;           }     }}ll powmod(ll x,ll n){    if(n==0) return 1%mod;    ll temp=powmod(x,n>>1);    temp=temp*temp%mod;    if(n&1) temp=temp*x%mod;    return temp;}int main(){    int n;    while(scanf("%d%d",&mod,&n)&&mod||n)    {        int c;        for(int i=1;i<=n;i++)        {            bool tag=false;            memset(a,0,sizeof(a));            scanf("%d",&c);            pr(mod-1);            for(int j=0;j<=cnt;j++)            {                if(powmod(c,a[j])==1)                {                    tag=true;                    break;                }            }            if(powmod(c,mod-1)!=1)  tag=false;            if(tag) printf("NO\n");            else printf("YES\n");        }    }    return 0;}

In the field of Cryptography, prime numbers play an important role. We are interested in a scheme called "Diffie-Hellman" key exchange which allows two communicating parties to exchange a secret key. This method requires a prime number p and r which is a primitive root of p to be publicly known. For a prime number p, r is a primitive root if and only if it's exponents r, r², r³, ... , r^p-1 are distinct (mod p).

Cryptography Experts Group (CEG) is trying to develop such a system. They want to have a list of prime numbers and their primitive roots. You are going to write a program to help them. Given a prime number p and another integer r < p , you need to tell whether r is a primitive root of p.

Input

There will be multiple test cases. Each test case starts with two integers p ( p < 2 ³¹ ) and n (1 ≤ n ≤ 100 ) separated by a space on a single line. p is the prime number we want to use and n is the number of candidates we need to check. Then n lines follow each containing a single integer to check. An empty line follows each test case and the end of test cases is indicated by p=0 and n=0 and it should not be processed. The number of test cases is atmost 60.

Output

For each test case print "YES" (quotes for clarity) if r is a primitive root of p and "NO" (again quotes for clarity) otherwise.

Example

Input:5 2347 2340 0Output:YESNOYESNO

Explanation

In the first test case 3¹, 3² , 3³ and 3⁴ are respectively 3, 4, 2 and 1 (mod 5). So, 3 is a primitive root of 5.

4¹, 4² , 4³ and 4⁴ are respectively 4, 1, 4 and 1 respectively. So, 4 is not a primitive root of 5.