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;
void m_simple(vector<long long> &a, long long n) {
  vector<bool> used(n, false);
  for (long long i = 1; i < n; i++) {
    if (!used[i]) {
      a.push_back(i + 1);
      for (long long j = i; j < n; j += i + 1) {
        used[j] = true;
      }
    }
  }
}
int main() {
  long long n, k;
  scanf("%I64d", &n);
  scanf("%I64d", &k);
  vector<long long> mults, nums(n);
  vector<long long> simples;
  m_simple(simples, 1000000);
  bool flag = false;
  for (long long i = 0; i < n; i++) {
    scanf("%I64d", &nums[i]);
  }
  for (long long i = 0; simples[i] * simples[i] <= k; i++) {
    if (k % simples[i] == 0) {
      long long cur = 1;
      while (k % simples[i] == 0) {
        cur *= simples[i];
        k /= simples[i];
      }
      mults.push_back(cur);
    }
  }
  if (k > 1) mults.push_back(k);
  for (long long i = 0; i < mults.size(); i++) {
    flag = false;
    for (long long j = 0; j < n; j++) {
      if (nums[j] % mults[i] == 0) {
        flag = true;
        break;
      }
    }
    if (!flag) {
      cout << "No";
      cin >> n;
      return 0;
    }
  }
  cout << "Yes";
  cin >> n;
  return 0;
}