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 tam = 10000;
const int MOD = 1e9 + 7;
const int MOD1 = 988244535;
const double EPS = 1e-9;
const double PI = acos(-1);
vector<int> pri, pric;
bitset<10000> bi;
void criba() {
  bi.set();
  for (int i = 2; i < 10000; i++) {
    if (bi[i]) {
      for (int j = i + i; j < 10000; j += i) bi[j] = 0;
      pri.push_back(i);
      pric.push_back(i * i);
    }
  }
}
int fac[2000000];
void fact(int n) {
  int co;
  for (int i = 0; pric[i] <= n; i++) {
    if (n % pri[i] == 0) {
      co = 0;
      while (n % pri[i] == 0) {
        n /= pri[i];
        co++;
      }
      fac[pri[i]] = max(fac[pri[i]], co);
    }
  }
  if (n > 1) fac[n] = max(fac[n], 1);
}
bool facto(int n) {
  int co;
  for (int i = 0; pric[i] <= n; i++) {
    if (n % pri[i] == 0) {
      co = 0;
      while (n % pri[i] == 0) {
        n /= pri[i];
        co++;
      }
      if (fac[pri[i]] < co) return 0;
    }
  }
  if (n > 1) {
    if (fac[n] == 0) return 0;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  ;
  int a;
  criba();
  while (n--) {
    cin >> a;
    fact(a);
  }
  cout << (facto(k) ? "Yes" : "No") << '\n';
  return 0;
}