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> pr;
int cnt[1000009];
bool vis[1009];
int main() {
  int a, b, c, d, e, i, j, k, l, n, m, x, y, t, p;
  a = 1000;
  b = sqrtl(a);
  pr.push_back(2);
  for (i = 3; i <= b; i += 2) {
    if (!vis[i]) {
      for (j = i * i; j <= a; j += (i + i)) {
        vis[j] = true;
      }
    }
  }
  for (i = 3; i <= a; i += 2) {
    if (!vis[i]) {
      pr.push_back(i);
    }
  }
  l = pr.size();
  scanf("%d%d", &n, &k);
  for (j = 0; j < n; j++) {
    scanf("%d", &a);
    for (i = 0; i < l; i++) {
      if (pr[i] * pr[i] > a) {
        break;
      }
      b = 0;
      while (a % pr[i] == 0) {
        a = a / pr[i];
        b++;
      }
      cnt[pr[i]] = max(cnt[pr[i]], b);
    }
    if (a != 1) {
      cnt[a] = max(cnt[a], (int)1);
    }
  }
  c = 1;
  for (i = 0; i < l; i++) {
    if (pr[i] * pr[i] > k) {
      break;
    }
    b = 0;
    while (k % pr[i] == 0) {
      b++;
      k = k / pr[i];
    }
    if (cnt[pr[i]] < b) {
      c = 0;
    }
  }
  if (k != 1) {
    b = 1;
    if (cnt[k] < b) {
      c = 0;
    }
  }
  if (c) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }
}