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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 1e6 + 10;
map<int, int> mpc, mpk;
bool p[N];
vector<pair<int, int> > fact[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 2; i < N; i++) {
    if (!p[i]) {
      fact[i].push_back({i, 1});
      for (int j = i + i; j < N; j += i) {
        p[j] = true;
        int k = j, cnt = 0;
        while (k % i == 0) {
          k /= i;
          cnt++;
        }
        fact[j].push_back({i, cnt});
      }
    }
  }
  int n, k;
  cin >> n >> k;
  for (auto it : fact[k]) mpk[it.first] = it.second;
  for (int i = 0; i < n; i++) {
    int c;
    cin >> c;
    for (auto it : fact[c]) {
      mpc[it.first] = max(mpc[it.first], it.second);
    }
  }
  for (auto it = mpk.begin(); it != mpk.end(); it++) {
    if (it->second > mpc[it->first]) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}