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[1000005];
vector<vector<pair<int, int> > > cfact(1000005);
vector<int> p;
void prime() {
  for (int(i) = (2); (i) < (1000005); i++) {
    int fail = 0;
    for (int(j) = (2); (j) < (int(sqrt(i)) + 1); j++) {
      if (i % j == 0) fail = 1;
      if (fail) break;
    }
    if (!fail) p.push_back(i);
  }
  cout << p.size() << endl;
}
int maxprime[1000005];
int factor(int k, int n) {
  for (int(i) = (0); (i) < (n); i++) {
    int copy = c[i];
    for (int j = 0; j < p.size() and copy > 1; j++) {
      int times = 0;
      while (copy % p[j] == 0) {
        copy = copy / p[j];
        times++;
      }
      if (times > 0) cfact[i].push_back(make_pair(p[j], times));
      if (copy <= 1) break;
    }
  }
  for (int(i) = (0); (i) < (1000005); i++) maxprime[i] = 0;
  for (int(i) = (0); (i) < (n); i++) {
    for (int(j) = (0); (j) < (cfact[i].size()); j++) {
      maxprime[cfact[i][j].first] =
          max(maxprime[cfact[i][j].first], cfact[i][j].second);
    }
  }
  int fail = 0;
  for (int i = 0; i < p.size(); i++) {
    int times = 0;
    while (k % p[i] == 0) {
      k = k / p[i];
      times++;
    }
    if (times > maxprime[p[i]]) fail = 1;
    if (fail) break;
    if (k <= 1) break;
  }
  return fail;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  long long mul = 1;
  int ret = 0;
  for (int(i) = (0); (i) < (n); i++) {
    scanf("%d", &c[i]);
    mul = (mul * (c[i])) / (gcd(c[i], mul));
    if (mul % k == 0) ret = 1;
    if (ret) break;
    mul = mul % k;
  }
  if (ret == 0)
    cout << "NO" << endl;
  else
    cout << "YES" << endl;
}