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 N = 1000020;
int lp[N + 1];
int h[N + 1];
vector<int> pr;
vector<pair<long long, long long> > get(int k) {
  vector<pair<long long, long long> > divs;
  while (k != 1) {
    int t = 0;
    int cur = lp[k];
    while (k % cur == 0) {
      k /= cur;
      t++;
    }
    divs.push_back(make_pair(cur, t));
  }
  return divs;
}
int R[1000100];
int main() {
  ios_base::sync_with_stdio(0);
  for (int i = 2; i <= N; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int j = 0; j < (int)pr.size() && pr[j] <= lp[i] && i * pr[j] <= N;
         ++j) {
      lp[i * pr[j]] = pr[j];
      h[i * pr[j]] = i;
    }
  }
  int n, k;
  cin >> n >> k;
  vector<pair<long long, long long> > divs = get(k);
  for (int i = 0; i < divs.size(); ++i) R[divs[i].first] = divs[i].second;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    divs = get(x);
    for (int j = 0; j < divs.size(); ++j)
      if (R[divs[j].first] <= divs[j].second) R[divs[j].first] = 0;
  }
  bool ok = true;
  for (int i = 0; i <= 1000000 && ok; ++i) ok = (ok && (R[i] == 0));
  if (ok)
    cout << "Yes";
  else
    cout << "No";
  return 0;
}