Problem: 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 <image>. There are n ancient numbers c1, c2, ..., cn and Pari has to tell Arya <image> 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 <image> for any positive integer x?

Note, that <image> 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 <image> because 5 is one of the ancient numbers.

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1111111;
int a[N];
int n, k;
vector<int> prime, p;
void Getprime() {
  for (int i = 2; i <= 1000; i++) {
    bool bo = 1;
    for (int j = 2; j * j <= i; j++) {
      if (i % j == 0) {
        bo = 0;
        break;
      }
    }
    if (bo) {
      prime.push_back(i);
    }
  }
}
int main() {
  Getprime();
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  int K = k;
  for (int i = 0; i < prime.size() && prime[i] * prime[i] <= k; ++i) {
    int t = prime[i];
    if (k % t == 0) {
      int m = 1;
      while (K % t == 0 && K) {
        m *= t;
        K /= t;
      }
      p.push_back(m);
    }
  }
  if (K != 1) {
    p.push_back(K);
  }
  for (int i = 0; i < p.size(); ++i) {
    int pr = p[i];
    bool flag = false;
    for (int j = 1; j <= n; ++j) {
      if (a[j] % pr == 0) {
        flag = true;
        break;
      }
    }
    if (!flag) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
}