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 n, k, m = 1000010, p[1010000], a[1010000], c[1000000];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &c[i]);
  if (k == 1) {
    printf("Yes");
    return 0;
  }
  for (int i = 2; i < m; i++) {
    if (p[i] == 0) {
      for (int j = i; j < m; j += i) p[j] = i;
    }
  }
  fill(a, a + m, 0);
  vector<pair<int, int> > t, tk;
  for (int i = 0; i < n; i++) {
    if (c[i] == 1) continue;
    int s = c[i];
    t.clear();
    t.push_back(make_pair(p[s], 1));
    s /= p[s];
    while (s > 1) {
      bool nn = 0;
      for (int j = 0; j < t.size(); j++) {
        if (p[s] == t[j].first) {
          t[j].second++;
          nn = 1;
        }
      }
      if (nn == 0) t.push_back(make_pair(p[s], 1));
      s /= p[s];
    }
    for (int j = 0; j < t.size(); j++) {
      int f = t[j].first;
      a[f] = max(a[f], t[j].second);
    }
  }
  tk.push_back(make_pair(p[k], 1));
  k /= p[k];
  while (k > 1) {
    bool nn = 0;
    for (int j = 0; j < tk.size(); j++) {
      if (p[k] == tk[j].first) {
        tk[j].second++;
        nn = 1;
      }
    }
    if (nn == 0) tk.push_back(make_pair(p[k], 1));
    k /= p[k];
  }
  for (int i = 0; i < tk.size(); i++) {
    int r = tk[i].first;
    if (tk[i].second > a[r]) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}