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;
long long _sieve_size;
int bs[10000010];
void sieve(long long upperbound) {
  _sieve_size = upperbound + 1;
  memset(bs, 0, sizeof(bs));
  ;
  bs[0] = bs[1] = 0;
  for (long long i = 2; i <= _sieve_size; i++)
    if (!bs[i]) {
      bs[i] = i;
      for (long long j = i * i; j <= _sieve_size; j += i) {
        if (!bs[j]) bs[j] = i;
      }
    }
}
void generateprimes(int x, map<int, int>& mm) {
  while (bs[x]) {
    mm[bs[x]]++;
    x /= bs[x];
  }
}
int totient(int x) {
  int result = x;
  while (bs[x]) {
    int z = bs[x];
    while (z == bs[x]) x /= z;
    result -= result / z;
  }
  return result;
}
map<int, int> mm;
void solve() {
  sieve(2e6);
  int n, k;
  scanf("%d %d", &n, &k);
  map<int, int> mk;
  generateprimes(k, mk);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    map<int, int> my;
    generateprimes(x, my);
    for (auto j = my.begin(); j != my.end(); j++) {
      int num = j->first;
      int freq = j->second;
      mm[num] = max(mm[num], freq);
    }
  }
  int ok = 1;
  for (auto i = mk.begin(); i != mk.end(); i++) {
    int num = i->first;
    int freq = i->second;
    if (mm[num] < freq) ok = 0;
  }
  if (ok) {
    printf("Yes");
  } else {
    printf("No");
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  cin.sync_with_stdio(0);
  cout.sync_with_stdio(0);
  solve();
  return 0;
}