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 int SIEVE_LIMIT = 1000001;
int lp[SIEVE_LIMIT + 1];
vector<int> pr;
void linear_sieve() {
  for (int i = 2; i <= SIEVE_LIMIT; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int j = 0;
         j < (int)pr.size() && pr[j] <= lp[i] && i * pr[j] <= SIEVE_LIMIT; ++j)
      lp[i * pr[j]] = pr[j];
  }
}
map<int, int> factors;
bool f[101];
int numbers[101], nn = 0;
vector<pair<int, int> > A;
int a;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  linear_sieve();
  while (k != 1) {
    int num = 1;
    int pr = lp[k];
    while (k % pr == 0) {
      num = num * pr;
      k = k / pr;
    }
    numbers[nn++] = num;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    for (int j = 0; j < nn; j++) {
      if (a % numbers[j] == 0) {
        f[j] = true;
      }
    }
  }
  int cc = 0;
  for (int i = 0; i < nn; i++) {
    cc += f[i];
  }
  if (cc == nn) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
  return 0;
}