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>
int prime[100001], isprime[1000001];
int cnt[1000001];
int main() {
  int top = 0;
  for (int i = 2; i <= 1000000; i++) {
    if (isprime[i] == 0) {
      prime[top++] = i;
      for (long long j = (long long)i * i; j <= 1000000; j = j + i) {
        isprime[j] = 1;
      }
    }
  }
  int n, k;
  scanf("%d%d", &n, &k);
  int c;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c);
    for (int j = 0; j < top && (long long)prime[j] * prime[j] <= c; j++) {
      if (c % prime[j] == 0) {
        int now = 0;
        while (c % prime[j] == 0) {
          now++;
          c = c / prime[j];
        }
        if (now > cnt[prime[j]]) cnt[prime[j]] = now;
      }
    }
    if (c != 1) {
      if (cnt[c] == 0) cnt[c] = 1;
    }
  }
  int ok = 1;
  for (int j = 0; j < top && (long long)prime[j] * prime[j] <= k; j++) {
    if (k % prime[j] == 0) {
      int now = 0;
      while (k % prime[j] == 0) {
        now++;
        k = k / prime[j];
      }
      if (now > cnt[prime[j]]) {
        ok = 0;
        break;
      }
    }
  }
  if (k != 1) {
    if (cnt[k] == 0) ok = 0;
  }
  if (ok)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}