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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long a[1000005], mark[1000005], fr[1000005], b[1000005];
vector<long long> v;
void sieve() {
  long long i, j;
  v.push_back(2);
  for (i = 3; i <= 1000000; i += 2) {
    if (mark[i] == 0) {
      v.push_back(i);
      for (j = i + i; j <= 1000000; j += i) mark[j] = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sieve();
  long long n, k, i, j, x;
  while (cin >> n >> k) {
    for (i = 1; i <= n; i++) {
      cin >> x;
      a[x] = 1;
    }
    for (i = 0; i < v.size(); i++) {
      for (j = v[i]; j <= 1000000; j += v[i]) {
        if (!a[j]) continue;
        long long cn = 0, x = j;
        while (x % v[i] == 0) {
          x /= v[i];
          cn++;
        }
        b[v[i]] = max(b[v[i]], cn);
      }
    }
    bool f = 0;
    for (i = 0; i < v.size(); i++) {
      long long cn = 0;
      while (k % v[i] == 0) {
        k /= v[i];
        cn++;
      }
      if (cn > b[v[i]]) f = 1;
    }
    if (f)
      cout << "NO" << endl;
    else
      cout << "YES" << endl;
  }
  return 0;
}