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 long long NR = 1e6 + 100;
const long long oo = 1e16 + 100;
const long long MOD = 1e9 + 7;
const long long MOD2 = 1e9 + 9;
int n, k;
int ciur[NR];
int a[NR];
vector<string> ret;
int mappp[NR];
int mapp[NR];
int sw = 1;
signed main() {
  ret.push_back("No");
  ret.push_back("Yes");
  long long i, j;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (i = 2; i <= 1000000; ++i) {
    if (!ciur[i]) {
      for (j = i; j <= 1000000; j += i) {
        ciur[j] = i;
      }
    }
  }
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    vector<int> temp;
    while (ciur[a[i]] && a[i] % ciur[a[i]] == 0) {
      temp.push_back(ciur[a[i]]);
      mapp[ciur[a[i]]] = 1 + mapp[ciur[a[i]]];
      mappp[ciur[a[i]]] = max(mappp[ciur[a[i]]], mapp[ciur[a[i]]]);
      a[i] /= ciur[a[i]];
    }
    while (temp.size()) {
      mapp[temp.back()] = 0;
      temp.pop_back();
    }
  }
  for (int d = 2; d * d <= k; ++d) {
    int cnt = 0;
    while (k % d == 0) {
      ++cnt;
      k /= d;
    }
    if (cnt > mappp[d]) {
      sw = 0;
    }
  }
  if (k != 1 && 1 > mappp[k]) {
    sw = 0;
  }
  cout << ret[sw] << '\n';
  return 0;
}