cf #360 div2 D-Remainders Game(中国剩余定理)

Today Pari and Arya are playing a game called Remainders.

Pari chooses two positive integer x and k, and tells Arya k but not x. Arya have to find the value $\text{[math]}$ . There are n ancient numbers c1, c2, ..., cn and Pari has to tell Arya $\text{[math]}$ if Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or not. Formally, is it true that Arya can understand the value $\text{[math]}$ for any positive integer x?

Note, that $\text{[math]}$ means the remainder of x after dividing it by y.

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000 000) — the number of ancient integers and value k that is chosen by Pari.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 1 000 000).

Output

Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.

Examples
input
4 5
2 3 5 12
output
Yes
input
2 7
2 3
output
No

Note

In the first sample, Arya can understand $\text{[math]}$ because 5 is one of the ancient numbers.

In the second sample, Arya can't be sure what $\text{[math]}$ is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.

题意：给定k和n，给出n个ci，表示你可以知道x%ci。问能否唯一确定x

思路：首先，根据剩余定理，如果我们想知道x%m等于多少，当且仅当我们知道x%m1,x%m2..x%mr分别等于多少，其中m1m2...mr=m，并且mi相互互质，即构成独立剩余系。令m的素数分解为m=p1^k1p2^k2...pr^kr，如果任意i，都有pi^ki的倍数出现在集合中，那么m就能被猜出来。即k是否存在于素数集合或者存在k的整数倍。

#include <bits/stdc++.h>
using namespace std;

long long gcd(long long a,long long b){
    if(b == 0)  return a;
    return gcd(b,a%b);
}
long long lcm(long long a,long long b){
    return a*b/gcd(a,b);
}

int main()
{
    int n,k;
    while(cin>>n>>k){
        int ans = 1;
        int temp;
        for(int i = 1;i <= n;i ++){
            scanf("%d",&temp);
            ans = gcd(k,lcm(ans,temp));
        }
        if(ans == k)    cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}