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 prime[100000];
int tmp[100000];
int myp[100000];
int num[100000];
int ans[100000];
void fprime() {
  int atmp = 0, j;
  for (int i = 2; i < 100000; i++) {
    if (!prime[i]) {
      tmp[atmp++] = i;
    }
    for (j = 0; j < atmp; j++) {
      if (i * tmp[j] >= 100000) break;
      prime[i * tmp[j]] = 1;
      if (i % tmp[j] == 0) break;
    }
  }
}
int main() {
  fprime();
  int al = 0;
  int n, k;
  scanf("%d%d", &n, &k);
  int tk = k;
  int i = 0;
  while (tk > 1) {
    int nn = 0;
    if (tmp[i] >= 1000) {
      myp[al] = tk;
      num[al++] = 1;
      break;
    }
    while (tk % tmp[i] == 0) {
      nn++;
      tk /= tmp[i];
    }
    if (nn) {
      myp[al] = tmp[i];
      num[al++] = nn;
    }
    i++;
  }
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    for (int j = 0; j < al; j++) {
      int nn = 0;
      while (t % myp[j] == 0) {
        nn++;
        t /= myp[j];
      }
      ans[j] = max(ans[j], nn);
    }
  }
  bool ok = 1;
  for (int i = 0; i < al; i++) {
    if (ans[i] < num[i]) {
      ok = 0;
      break;
    }
  }
  if (ok)
    puts("Yes");
  else
    puts("No");
}