CodeForces-668D：Remainders Game （中国剩余定理||理解）

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 . There are n ancient numbers c₁, c₂, ..., c_n and Pari has to tell Arya  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  for any positive integer x?

Note, that  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, ..., c_n (1 ≤ c_i ≤ 1 000 000).

Output

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

Example

Input

4 5
2 3 5 12

Output

Yes

Input

2 7
2 3

Output

No

题意：给定N，K。输入N个ni，表示已知X%ni的值，有了N组这样的剩余系，问是否X%K的值唯一。

思路：求出Lcm(ni)，若Lcm(ni)%K==0，则唯一。

原因：因为中国剩余定理：ans=Σ（Ai*Ni*Mi）%N，而N就是Lcm(ni)。如果N是K的倍数，那么先%N，再%K的结果是不变的。

 

 

#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
ll gcd(ll a,ll b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}
int main()
{
    ll N,K,x,res=1;
    scanf("%lld%lld",&N,&K);
    for(int i=1;i<=N;i++){
        scanf("%lld",&x);
        res=x/gcd(x,res)*res%K;
    }
    if(res%K==0) printf("Yes\n");
    else printf("No\n");
    return 0;
}