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;
int c[1000010];
bool notP[1000010];
int primes[1000010];
int sz = 0;
int cnt[1000010];
int main() {
  for (int i = 2; i <= 1000; i++) {
    for (int j = i + i; j <= 1000000; j += i) {
      notP[j] = 1;
    }
  }
  for (int i = 2; i <= 1000; i++) {
    if (!notP[i]) {
      primes[sz++] = i;
    }
  }
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < sz; j++) {
      int cur = 0;
      while (c[i] % primes[j] == 0) {
        cur++;
        c[i] /= primes[j];
      }
      cnt[primes[j]] = max(cnt[primes[j]], cur);
    }
    cnt[c[i]] = 1;
  }
  bool ok = 1;
  for (int i = 0; i < sz; i++) {
    int cur = 0;
    while (k % primes[i] == 0) {
      cur++;
      k /= primes[i];
    }
    if (cnt[primes[i]] < cur) {
      ok = 0;
    }
  }
  if (cnt[k] < 1) {
    ok = 0;
  }
  if (ok) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
  return 0;
}