cf----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 . There are n ancient numbersc₁, 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.

Sample Input

Input

4 5
2 3 5 12

Output

Yes

Input

2 7
2 3

Output

No

Hint

In the first sample, Arya can understand  because 5 is one of the ancient numbers.

In the second sample, Arya can't be sure what  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.

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m1*m2...*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^k1*p2^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。
这个问题等价于问LCM(ci)%m是否等于0
所以只要求出LCM(ci)即可，不过要边求lcm，边和m取gcd，防止爆int

ac代码：

#include<cstdio>  

#include<cstring>  

#include<iostream>  

#include<algorithm>  

#include<queue>  

#include<vector>  

using namespace std;  

const int maxn = 1e5+100;  

typedef long long LL;  

LL gcd(LL a,LL b) { return b==0 ? a:gcd(b,a%b);}  

LL lcm(LL a,LL b){ return a/gcd(a,b)*b; }  

int main()  

{  

int n,k,a,i; 

scanf("%d%d",&n,&k);  

LL ans=1;  

for(i=1;i<=n;i++)  

{  

scanf("%d",&a);  

ans=lcm(ans,a)%k;  

}  

printf("%s\n",ans==0 ? "Yes":"No");  

return 0;  

}