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 er[MAXN], c[MAXN], smpl[MAXN];
vector<pair<int, int> > dec(int x) {
  vector<pair<int, int> > ans;
  while (x > 1) {
    int cnt = 0;
    int dv = er[x];
    while (x % dv == 0) {
      ++cnt;
      x /= dv;
    }
    ans.push_back(make_pair(dv, cnt));
  }
  return ans;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
  }
  for (int i = 2; i < MAXN; ++i) {
    if (er[i] != 0)
      continue;
    else {
      er[i] = i;
      for (long long j = ((long long)i) * i; j < MAXN; j += i) {
        er[j] = i;
      }
    }
  }
  vector<pair<int, int> > tmp;
  for (int i = 0; i < n; ++i) {
    tmp = dec(c[i]);
    for (int j = 0; j < (int)tmp.size(); ++j) {
      smpl[tmp[j].first] = max(smpl[tmp[j].first], tmp[j].second);
    }
  }
  tmp = dec(k);
  for (int i = 0; i < (int)tmp.size(); ++i) {
    if (smpl[tmp[i].first] < tmp[i].second) {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}