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 double pi = 3.14159265358979323846;
const long double eps = 1e-12;
const unsigned long long mod = 1e9 + 7;
const int inf = INT_MAX;
const int size = 1000001;
vector<int> used(size, 0);
vector<int> fact_k(size, 0), fact_lcm(size, 0);
void Eratos(int n) {
  int p = 2;
  used[p] = 2;
  while (1) {
    int k = 2;
    while (k * p <= n) {
      used[k * p] = p;
      ++k;
    }
    int dp = p;
    for (long long i = p; i < n + 1; ++i) {
      if (!used[i]) {
        p = i;
        used[p] = p;
        break;
      }
    }
    if (dp == p) break;
  }
}
void Fact(vector<int>& f, int n) {
  int p = -1;
  while (n > 1) {
    int div = used[n];
    int count = 0;
    while (!(n % div)) {
      ++count;
      n /= div;
    }
    fact_k[div] = count;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  int n, k;
  cin >> n >> k;
  Eratos(size);
  Fact(fact_k, k);
  for (long long i = 0; i < n; ++i) {
    int c;
    cin >> c;
    int p = -1;
    while (c > 1) {
      int div = used[c];
      int count = 0;
      while (!(c % div)) {
        ++count;
        c /= div;
      }
      fact_lcm[div] = max(fact_lcm[div], count);
    }
  }
  bool ans = true;
  for (long long i = 0; i < size; ++i) {
    if (fact_lcm[i] < fact_k[i]) ans = false;
  }
  if (!ans) {
    cout << "No";
  } else {
    cout << "Yes";
  }
  return 0;
}