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 maxn = 1e6 + 10;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
long long n, k, a[maxn], isnp[maxn], isv[maxn], t, cnt[maxn], p;
bool flag;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  isv[1] = 1;
  for (int i = 2; i < maxn; i++) {
    if (!isnp[i]) {
      isv[i] = i;
      for (int j = 2 * i; j < maxn; j += i) {
        if (isv[j] == 0) isv[j] = i;
        isnp[j] = 1;
      }
    }
  }
  for (int i = 0; i != n; i++) {
    cin >> a[i];
    while (a[i] > 1) {
      p = isv[a[i]];
      t = 0;
      while (a[i] % p == 0) {
        a[i] /= p;
        t++;
      }
      cnt[p] = max(cnt[p], t);
    }
  }
  while (k > 1) {
    p = isv[k];
    t = 0;
    while (k % p == 0) {
      k /= p;
      t++;
    }
    if (cnt[p] < t) flag = 1;
  }
  if (flag == 1)
    cout << "No";
  else
    cout << "Yes";
  return 0;
}