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;
bool ok = true;
int cntr[1000000];
bool vis[1000000];
bool notPrime[1000000];
inline void initPrimes() {
  notPrime[0] = notPrime[1] = true;
  for (int i = 2; i < 1000000; i += 1 + (i & 1)) {
    if (!notPrime[i]) {
      for (int j = i + i; j < 1000000; j += i) {
        notPrime[j] = true;
      }
    }
  }
}
inline void factorize(int num, void (*fu)(int, int)) {
  for (int i = 2; i * i <= num && ok; i += 1 + (i & 1)) {
    if (!notPrime[i] && num % i == 0) {
      int cntr = 0;
      while (num % i == 0) {
        ++cntr;
        num /= i;
      }
      fu(i, cntr);
    }
  }
  if (num > 1) {
    fu(num, 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  initPrimes();
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    int num;
    cin >> num;
    if (!vis[num]) {
      factorize(
          num, [](int prime, int cnt) { cntr[prime] = max(cntr[prime], cnt); });
      vis[num] = true;
    }
  }
  factorize(k, [](int prime, int cnt) {
    if (cntr[prime] < cnt) {
      ok = false;
    }
  });
  cout << (ok ? "Yes\n" : "No\n");
  return 0;
}