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 po(int a, int b) {
  int re = 1;
  while (b) {
    if (b % 2) re *= a;
    a *= a;
    b /= 2;
  }
  return re;
}
int prime[1000010], c[1000010];
vector<int> co;
int main() {
  ios_base::sync_with_stdio(0);
  int n, k, p, f, temp;
  cin >> n >> k;
  for (int i = 2; i <= sqrt(1000000); i++) {
    if (!prime[i])
      for (int j = i * i; j <= 1000000; j += i) prime[j] = 1;
  }
  for (int i = 2; i <= 1000000; i++)
    if (!prime[i]) {
      temp = 1;
      while (k != 1 && k % i == 0) {
        k /= i;
        temp *= i;
      }
      co.push_back(temp);
      if (k == 1) break;
    }
  for (int i = 0; i < n; i++) cin >> c[i];
  for (int i = 0; i < co.size(); i++)
    if (co[i] > 1) {
      f = 0;
      for (int j = 0; j < n; j++)
        if (c[j] % co[i] == 0) {
          f = 1;
          break;
        }
      if (!f) {
        cout << "No\n";
        return 0;
      }
    }
  cout << "Yes\n";
  return 0;
}