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 gcd(int a, int b) {
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
int n, k;
int vis[1000010];
int prime[100010], cou;
void get_prime() {
  for (int i = 2; i <= 1000000; i++) {
    if (!vis[i]) {
      prime[cou++] = i;
    }
    for (int j = 0; j < cou && i * prime[j] <= 1000000; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        break;
      }
    }
  }
}
vector<pair<int, int> > v;
int main() {
  get_prime();
  int a;
  scanf("%d%d", &n, &k);
  int g = k;
  pair<int, int> temp;
  for (int i = 0; i < cou && k >= prime[i]; i++) {
    if (k % prime[i] == 0) {
      temp.first = prime[i];
      temp.second = 0;
      while (k % prime[i] == 0) {
        k /= prime[i];
        temp.second++;
      }
      v.push_back(temp);
    }
  }
  int size = v.size();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    for (int j = 0; j < size && a >= v[j].first; j++) {
      if (v[j].second > 0 && a % v[j].first == 0) {
        int cnt = 0, re = 1;
        while (cnt < v[j].second && a % v[j].first == 0) {
          cnt++;
          a /= v[j].first;
          re *= v[j].first;
        }
        if (cnt >= v[j].second) {
          v[j].second = 0;
          g /= re;
        }
      }
    }
  }
  if (g != 1) {
    puts("No");
  } else {
    puts("Yes");
  }
  return 0;
}