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;
const int max_n = 1e6 + 5;
map<int, int> primeslcm;
int c[max_n];
int lp[max_n];
vector<int> pr;
void fact(int n) {
  while (n > 1) {
    int lea = lp[n];
    int exp = 0;
    while (n % lea == 0) {
      exp++;
      n /= lea;
    }
    if (primeslcm.find(lea) == primeslcm.end()) {
      primeslcm.insert({lea, exp});
    } else {
      primeslcm[lea] = max(primeslcm[lea], exp);
    }
  }
}
void sieve() {
  for (long long int i = 2; i < max_n; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (long long int j = 0;
         j < pr.size() && pr[j] <= lp[i] && i * pr[j] < max_n; j++) {
      lp[i * pr[j]] = pr[j];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  sieve();
  int n;
  long long int k;
  cin >> n >> k;
  long long int lcm = 1;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    fact(c[i]);
  }
  while (k > 1) {
    int lea = lp[k];
    int exp = 0;
    while (k % lea == 0) {
      exp++;
      k /= lea;
    }
    if (primeslcm.find(lea) == primeslcm.end() || primeslcm[lea] < exp) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}