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];
pair<int, int> p[10005];
int ans[10005];
int prime[600001], cnt = 0;
bool flag[1000005];
void init() {
  for (int i = 2; i <= 1000000; i++) {
    if (!flag[i]) {
      prime[cnt++] = i;
      flag[i] = 1;
    }
    for (int j = 0; j < cnt && i * prime[j] <= 1000000; j++) {
      flag[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        break;
      }
    }
  }
}
int main() {
  int n, k, i, j, m = 0;
  cin >> n >> k;
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  init();
  int tmp = k;
  for (i = 0; i < cnt; i++) {
    if (tmp % prime[i] == 0) {
      p[m].first = prime[i];
      while (tmp % prime[i] == 0) {
        p[m].second++;
        tmp /= prime[i];
      }
      m++;
    }
  }
  if (tmp != 1) {
    p[m].first = tmp;
    p[m++].second = 1;
  }
  for (i = 1; i <= n; i++) {
    if (c[i] % k == 0) {
      printf("Yes\n");
      return 0;
    }
    for (j = 0; j < m; j++) {
      int sum = 0;
      if (c[i] % p[j].first == 0) {
        while (c[i] % p[j].first == 0) {
          sum++;
          c[i] /= p[j].first;
        }
      }
      ans[j] = max(ans[j], sum);
    }
  }
  int fail = 0;
  for (i = 0; i < m; i++) {
    if (ans[i] < p[i].second) {
      fail = 1;
      break;
    }
  }
  if (fail)
    printf("No\n");
  else
    printf("Yes\n");
}