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;
int c[1000005];
int main() {
  int i, j, k;
  int n;
  scanf("%d", &n);
  scanf("%d", &k);
  for (i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
  }
  int x = sqrt(k);
  for (i = 2; i <= x; i++) {
    int temp = 0;
    while (k % i == 0) {
      temp++;
      k /= i;
    }
    if (temp != 0) {
      int temp3 = 0;
      for (j = 1; j <= n; j++) {
        int temp2 = 0;
        while (c[j] % i == 0) {
          c[j] /= i;
          temp2++;
        }
        temp3 = max(temp3, temp2);
      }
      if (temp > temp3) {
        printf("No\n");
        return 0;
      }
    }
  }
  if (k != 1) {
    int temp = 1;
    if (temp != 0) {
      int temp3 = 0;
      for (j = 1; j <= n; j++) {
        int temp2 = 0;
        while (c[j] % k == 0) {
          c[j] /= k;
          temp2++;
        }
        temp3 = max(temp3, temp2);
      }
      if (temp > temp3) {
        printf("No\n");
        return 0;
      }
    }
  }
  printf("Yes\n");
  return 0;
}