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 MOD = 1e9 + 7;
const int INF = 1e9;
const int N = 1e6 + 5;
vector<int> primes;
void trival_prime(int r) {
  for (int i = (2); i < (r + 1); i++) {
    bool ok = true;
    for (auto d : primes) {
      if (d * d > i) break;
      if (i % d == 0) {
        ok = false;
        break;
      }
    }
    if (ok) primes.push_back(i);
  }
}
vector<int> factor(int n) {
  vector<int> ans;
  for (auto d : primes) {
    if (d * d > n) break;
    while (n % d == 0) {
      ans.push_back(d);
      n /= d;
    }
  }
  if (n > 1) ans.push_back(n);
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> c(n);
  for (int i = (0); i < (n); i++) cin >> c[i];
  map<int, int> cfact, kfact;
  trival_prime(N);
  vector<int> kf = factor(k);
  for (auto d : kf) kfact[d]++;
  bool ok = true;
  for (auto [prime, times] : kfact) {
    int v = 1;
    for (int i = (0); i < (times); i++) v *= prime;
    bool flag = false;
    for (int i = (0); i < (n); i++) {
      if (c[i] % v == 0) {
        flag = true;
        break;
      }
    }
    if (!flag) {
      ok = false;
      break;
    }
  }
  if (ok)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}