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 dv[1010101];
int has[1010101];
int ma[1010101];
int h[1010101];
void add(long long x, int t) {
  while (x > 1) {
    has[dv[x]] += t;
    x /= dv[x];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (long long i = 2; i <= 1000000; i++) {
    if (dv[i] == 0) {
      for (long long j = 1; j * i <= 1000000; j++) {
        dv[i * j] = i;
      }
    }
  }
  long long n, k;
  cin >> n >> k;
  add(k, 1);
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    h[x] = 1;
  }
  for (int i = 2; i <= 1000000; i++) {
    if (h[i]) {
      int ed = 0;
      int c = 0;
      int x = i;
      while (x > 1) {
        if (dv[x] != ed) {
          c = 0;
        }
        c++;
        ma[dv[x]] = max(ma[dv[x]], c);
        ed = dv[x];
        x /= dv[x];
      }
    }
  }
  for (int i = 2; i <= 1000000; i++) {
    if (has[i] > ma[i]) {
      cout << "No\n";
      return 0;
    }
  }
  cout << "Yes" << endl;
}