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;
long long prime[1000009], T, keep[1000009], dp[1000009];
bool isp[1000009];
vector<long long> NOW;
void precalc() {
  long long i, j;
  T = 0;
  for (i = 2; i <= 1000000; i++) {
    if (isp[i] == false) {
      for (j = i * i; j <= 1000000; j += i) {
        isp[j] = true;
      }
      prime[T] = i;
      T++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long i, j, k, l, n, m, temp, ii, siz;
  precalc();
  cin >> n >> m;
  for (i = 0; prime[i] * prime[i] <= m && i < T; i++) {
    temp = 0;
    while (m % prime[i] == 0) {
      temp++;
      m /= prime[i];
    }
    keep[prime[i]] = temp;
    if (temp) NOW.push_back(prime[i]);
  }
  if (m > 1) {
    keep[m] = 1;
    NOW.push_back(m);
  }
  siz = NOW.size();
  for (ii = 1; ii <= n; ii++) {
    cin >> m;
    for (i = 0; i < siz; i++) {
      temp = 0;
      while (m % NOW[i] == 0) {
        temp++;
        m /= NOW[i];
      }
      dp[NOW[i]] = max(dp[NOW[i]], temp);
    }
  }
  for (i = 0; i < T; i++) {
    if (keep[prime[i]] > dp[prime[i]]) {
      cout << "No\n";
      return 0;
    }
  }
  cout << "Yes\n";
}