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 long long mod = 1e9 + 7, siz = 1e6 + 5;
long long t, n, m, k, prime[siz], factors[siz];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (long long i = 2; i < siz; i++) {
    if (prime[i]) continue;
    for (long long j = i; j < siz; j += i) prime[j] = i;
  }
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    while (x > 1) {
      long long cur = prime[x], cnt = 0;
      while (x % cur == 0) cnt++, x /= cur;
      factors[cur] = max(factors[cur], cnt);
    }
  }
  long long f = 1;
  while (k > 1) {
    f &= (factors[prime[k]] > 0);
    factors[prime[k]]--;
    k /= prime[k];
  }
  if (f)
    cout << "Yes\n";
  else
    cout << "No\n";
}