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;
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int res = power(a, b / 2, m);
  if (b % 2) {
    return (((res * res) % m) * a) % m;
  }
  return res * res % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  const int N = 1000000;
  vector<int> lp(N + 1, 0);
  vector<int> pr;
  for (int i = 2; i <= N; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int j = 0; j < (int)pr.size() && pr[j] <= lp[i] && i * pr[j] <= N; ++j)
      lp[i * pr[j]] = pr[j];
  }
  int n, k;
  cin >> n >> k;
  vector<int> c(n);
  for (int i = 0; i < n; i++) {
    cin >> c[i];
  }
  map<int, int> a;
  int l = k;
  while (l != 1) {
    a[lp[l]]++;
    l /= lp[l];
  }
  int is = true;
  for (int i = 0; i < n; i++) {
    is = true;
    map<int, int> b;
    int j = c[i];
    while (j != 1) {
      b[lp[j]]++;
      j /= lp[j];
    }
    for (auto o : b) {
      if (a[o.first] <= b[o.first]) {
        k /= power(o.first, a[o.first], 10000000007);
        a[o.first] = 0;
      }
    }
    if (k == 1) break;
  }
  if (k == 1) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}