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;
vector<int> divisors(int n) {
  vector<int> v;
  int i = 2;
  while (i * i <= n)
    if (n % i == 0) {
      n /= i;
      v.push_back(i);
    } else
      i++;
  if (n != 1) v.push_back(n);
  return v;
}
const int maxN = (int)2e6;
int divs[maxN] = {};
int pDiv[maxN] = {};
int wDiv[maxN] = {};
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  vector<int> v(n);
  for (int i = 0; i < n; i++) scanf("%d", &v[i]);
  vector<int> d = divisors(k);
  vector<int> q;
  for (int i = 0; i < d.size(); i++) {
    if (i == 0 || d[i] != d[i - 1]) q.push_back(d[i]);
    divs[d[i]]++;
  }
  int qSz = q.size();
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < qSz; j++) pDiv[q[j]] = 0;
    int num = v[i];
    for (int j = 0; j < qSz; j++)
      while (num % q[j] == 0) num /= q[j], pDiv[q[j]]++;
    for (int j = 0; j < qSz; j++) wDiv[q[j]] = max(wDiv[q[j]], pDiv[q[j]]);
  }
  for (int i = 0; i < qSz; i++)
    if (wDiv[q[i]] < divs[q[i]]) {
      printf("No");
      return 0;
    }
  printf("Yes");
  return 0;
}