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 MAXN = 1000010;
int n, k, p[MAXN], cnt[MAXN], bio[MAXN], trebam[MAXN], imam[MAXN], divi[MAXN];
int main() {
  scanf("%d %d", &n, &k);
  for (int i = (0); i < int(n); i++) scanf("%d", &p[i]);
  for (int i = (2); i < int(MAXN); i++) {
    if (divi[i]) continue;
    for (int j = 1; j * i < MAXN; j++) divi[i * j] = i;
  }
  int K = k;
  vector<int> v;
  while (k > 1) {
    trebam[divi[k]]++;
    v.push_back(divi[k]);
    k /= divi[k];
  }
  sort((v).begin(), (v).end());
  v.erase(unique((v).begin(), (v).end()), v.end());
  for (int i = (0); i < int(n); i++) {
    int t = p[i];
    bool ok = 1;
    while (t > 1) {
      if (bio[divi[t]] != i + 1) {
        bio[divi[t]] = i + 1;
        cnt[divi[t]] = 1;
      } else
        cnt[divi[t]]++;
      t /= divi[t];
    }
    for (int j = (0); j < int((int)v.size()); j++)
      if (cnt[v[j]] >= trebam[v[j]]) imam[v[j]] = 1;
  }
  bool ok = 1;
  for (int i = (0); i < int((int)v.size()); i++)
    if (!imam[v[i]]) ok = 0;
  if (ok)
    puts("Yes");
  else
    puts("No");
  return 0;
}