CodeForces 687B

B. Remainders Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., cn 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, ..., 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 52 3 5 12

output

Yes

input

2 72 3

output

No

Note

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.

题目大意：

已知一个k和n个数c1,c2,c3...cn,问在已知x%c1,x%c2...x%cn的情况下，能否使x%k的值唯一。

思路：

中国剩余定理的转化，如果我知道x%c1,x%c2...x%cn我可以求出满足条件的x值，也可以得出其他的x（x = x+y*lcm(),y∈Z），但是如果要想结果唯一的话，即x%k == (x+lcm)%k;也就是a1*k+ z = x;a2*k + z = x + lcm;(a2 - a1)*k = lcm。所以只需要k是lcm约数就可以了，注意最小公倍数可能会超longlong，所以要一边算lcm一边和k求gcd，防止数据溢出。
附上代码

#include<cstdio>using namespace std;typedef long long ll;ll gcd(ll a,ll b){    return a == 0?b:gcd(b%a,a);}ll lcm(ll a,ll b){    return a*b/gcd(a,b);}int main(){    int n,k,num;    ll ans = 1;    int ok = 0;    scanf("%d%d",&n,&k);    for(int i = 0;i < n;i++)    {        scanf("%d",&num);        ans = lcm(ans,num);        ans = gcd(ans,k);        if(ans == k)            ok = 1;    }    if(ok)        printf("Yes\n");    else        printf("No\n");    return 0;}