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;
int pf[1000005], n, val, a[1000005], b[1000005], c[1000005];
vector<int> v;
void fact(int num) {
  while (pf[num] != -1) {
    c[pf[num]]++;
    v.push_back(pf[num]);
    num /= pf[num];
  }
  return;
}
int main() {
  memset(pf, -1, sizeof(pf));
  for (int i = (2); i <= (1000000); i++)
    if (pf[i] == -1) {
      for (int j = i; j <= 1000000; j += i)
        if (pf[j] == -1) pf[j] = i;
    }
  scanf("%d %d", &n, &val);
  fact(val);
  for (int i = (0); i <= ((int)v.size() - 1); i++) a[v[i]] = c[v[i]];
  for (int i = (0); i <= ((int)v.size() - 1); i++) c[v[i]] = 0;
  for (int i = (1); i <= (n); i++) {
    int x;
    scanf("%d", &x);
    v.clear();
    fact(x);
    for (int i = (0); i <= ((int)v.size() - 1); i++)
      b[v[i]] = max(b[v[i]], c[v[i]]), c[v[i]] = 0;
  }
  for (int i = (1); i <= (1000000); i++)
    if (a[i] > b[i]) {
      printf("No\n");
      return 0;
    }
  printf("Yes\n");
  return 0;
}