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 INF = numeric_limits<int>::max();
const long long LLINF = numeric_limits<long long>::max();
const unsigned long long ULLINF = numeric_limits<unsigned long long>::max();
const double PI = acos(-1.0);
int n, k;
int prime[1000010];
const int maxn = 1000010;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 2; i < maxn; i++) {
    if (!prime[i]) {
      prime[i] = i;
      for (int j = i * 2; j < maxn; j += i) prime[j] = i;
    }
  }
  map<int, int> kd, cd;
  while (k > 1) {
    kd[prime[k]]++;
    k /= prime[k];
  }
  for (int i = 0; i < n; i++) {
    int c;
    scanf("%d", &c);
    map<int, int> cur;
    while (c > 1) {
      cur[prime[c]]++;
      c /= prime[c];
    }
    for (auto K : cur) cd[K.first] = max(cd[K.first], K.second);
  }
  for (auto K : kd) {
    if (K.second > cd[K.first]) {
      printf("No");
      return 0;
    }
  }
  printf("Yes");
  return 0;
}