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>
const int MX = 2e6 + 7;
const int mod = 998244353;
const double pi = 3.1415926535897932384;
double isp = 1e-13;
using namespace std;
long long qpow(long long a, long long b, long long MOD = mod) {
  for (long long ans = 1;; a = a * a % MOD, b >>= 1) {
    if (b & 1) ans = ans * a % MOD;
    if (!b) return ans;
  }
}
long long inv(long long a, long long MOD = mod) {
  return qpow(a, MOD - 2, MOD);
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long ret = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return ret;
}
long long getInv(int a, int mod) {
  long long x, y;
  long long d = exgcd(a, mod, x, y);
  return d == 1 ? (x % mod + mod) % mod : -1;
}
int a[MX];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> mp;
  for (int i = 2; i <= k; i++) {
    if (k % i == 0) {
      int cnt = 0;
      while (k % i == 0) {
        k /= i;
        cnt++;
      }
      mp.push_back({i, cnt});
    }
  }
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    for (auto j : mp) {
      if (x % j.first == 0) {
        int cnt = 0;
        while (x % j.first == 0) {
          cnt++;
          x /= j.first;
        }
        a[j.first] = max(a[j.first], cnt);
      }
    }
  }
  for (auto x : mp) {
    if (x.second > a[x.first]) {
      cout << "No" << '\n';
      return 0;
    }
  }
  cout << "Yes" << '\n';
  return 0;
}