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 INF = 1e9 + 250;
long long mod = 1e9 + 7;
const int N = 1000005;
int a[N], b[N], f[N];
const int p_N = 1000000;
int not_p[p_N + 5], prime[p_N + 5], mpp[p_N + 5];
int pk = 0;
void get_p() {
  not_p[1] = 1;
  for (int i = 2; i < p_N; i++) {
    if (!not_p[i]) {
      mpp[i] = pk;
      prime[pk++] = i;
    }
    for (int j = 0; i * prime[j] < p_N; j++) {
      not_p[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  get_p();
  int n, m, x;
  int ok = 1;
  scanf("%d%d", &n, &m);
  for (int i = 0; m > 1; i++) {
    if (!not_p[m]) {
      a[m]++;
      break;
    }
    while (m % prime[i] == 0) {
      m = m / prime[i];
      a[prime[i]]++;
    }
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    if (f[x] == 1) continue;
    for (int j = 0; x > 1; j++) {
      if (!not_p[x]) {
        b[x] = max(b[x], 1);
        break;
      }
      int t = 0;
      while (x % prime[j] == 0) {
        x /= prime[j];
        t++;
      }
      b[prime[j]] = max(b[prime[j]], t);
    }
    f[x] = 1;
  }
  for (int i = 2; i <= 1000000; i++) {
    if (a[i] > b[i]) {
      ok = 0;
      break;
    }
  }
  puts(ok ? "Yes" : "No");
  return 0;
}