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 N = 1006;
int tot, p[N], vis[N];
void prime() {
  for (int i = 2; i < N; i++) {
    if (!vis[i]) p[tot++] = i;
    for (int j = 0; j < tot && i * p[j] < N; j++) {
      vis[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  prime();
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> factors;
  for (int i = 0; i < tot && p[i] * p[i] <= k; i++) {
    if (k % p[i] == 0) {
      int cnt = 0;
      while (k % p[i] == 0) {
        cnt++;
        k /= p[i];
      }
      factors.push_back({p[i], cnt});
    }
  }
  if (k > 1) {
    factors.push_back({k, 1});
  }
  for (int i = int(0); i <= int((n)-1); i++) {
    int x;
    cin >> x;
    for (int i = 0; i < factors.size(); i++) {
      if (factors[i].second == 0) continue;
      int p = factors[i].first;
      int cnt = 0;
      while (x % p == 0) {
        cnt++;
        x /= p;
      }
      if (cnt >= factors[i].second) factors[i].second = 0;
    }
  }
  for (int i = int(0); i <= int((factors.size()) - 1); i++)
    if (factors[i].second) {
      cout << "No" << endl;
      return 0;
    }
  cout << "Yes" << endl;
  return 0;
}