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 maxn = 1000001;
int n, k, c[maxn], m[maxn][10], mxx = 0, f[maxn][10], g[maxn], mx[maxn],
                                num[maxn];
bool have[maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    if (c[i] > mxx) mxx = c[i];
  }
  if (k > mxx) mxx = k;
  memset(f, 0, sizeof(f));
  for (int i = 2; i <= mxx; i++)
    if (!g[i])
      for (int j = i; j <= mxx; j += i) {
        g[j]++;
        int p = j;
        m[j][g[j]] = i;
        while (!(p % i)) p /= i, f[j][g[j]]++;
      }
  memset(have, 0, sizeof(have));
  for (int i = 1; i <= g[k]; i++) have[m[k][i]] = true, num[m[k][i]] = f[k][i];
  for (int i = 1; i <= n; i++)
    if (!(c[i] % k)) {
      printf("Yes\n");
      return 0;
    }
  memset(mx, 0, sizeof(mx));
  for (int i = 1; i <= n; i++) {
    int x = c[i];
    for (int j = 1; j <= g[x]; j++)
      if (have[m[x][j]]) mx[m[x][j]] = max(mx[m[x][j]], f[x][j]);
  }
  for (int i = 1; i <= g[k]; i++)
    if (mx[m[k][i]] < num[m[k][i]]) {
      printf("No\n");
      return 0;
    }
  printf("Yes\n");
  return 0;
}