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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
int prime[1000010];
int noprime[1000010];
int devide[1000010];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int i, j;
  memset(noprime, 0, sizeof(noprime));
  int size = 0;
  for (i = 2; i <= k; i++) {
    if (noprime[i] == 0) {
      prime[size++] = i;
      for (j = i + i; j <= k; j += i) {
        noprime[j] = 1;
      }
    }
  }
  int s1 = 0;
  for (i = 0; i < size; i++) {
    if (k % prime[i] == 0) {
      devide[s1] = prime[i];
      k /= prime[i];
      while (k % prime[i] == 0) {
        devide[s1] *= prime[i];
        k /= prime[i];
      }
      s1++;
    }
  }
  for (i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
    for (j = 0; j < s1; j++) {
      if (t % devide[j] == 0) {
        devide[j] = 1;
      }
    }
  }
  for (i = 0; i < s1; i++) {
    if (devide[i] != 1) {
      break;
    }
  }
  if (i >= s1) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }
}