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 larg = pow(10, 9) + 7;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long arr[], long long n, long long k) {
  long long curr = 1;
  long long i, dum, tmp;
  for (i = 0; i < n; i++) {
    dum = gcd(curr, arr[i]);
    tmp = (curr * arr[i]) / dum;
    curr = gcd(k, tmp);
  }
  return curr;
}
void solve(long long t) {
  long long n, k, lc, i;
  cin >> n >> k;
  long long arr[n];
  for (i = 0; i < n; i++) cin >> arr[i];
  ;
  lc = lcm(arr, n, k);
  if (lc == k)
    cout << "Yes\n";
  else
    cout << "No\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve(i);
  }
  return 0;
}