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>
#pragma GCC optimize("O3")
using namespace std;
enum { MOD0 = 1000000000, MOD7 = 1000000007, MOD9 = 1000000009 };
vector<int> cur;
vector<int> p;
vector<int> cnt;
vector<int> dd;
void factor(int x) {
  for (int i = 2; i * i <= x; ++i) {
    while (x % i == 0) {
      cur.push_back(i);
      x /= i;
    }
  }
  if (x != 1) {
    cur.push_back(x);
  }
  p.push_back(cur[0]);
  cnt.push_back(1);
  for (int i = 1; i < cur.size(); ++i) {
    if (cur[i] == cur[i - 1]) {
      cnt[cnt.size() - 1]++;
    } else {
      p.push_back(cur[i]);
      cnt.push_back(1);
    }
  }
  dd.resize(p.size(), 0);
}
void f(int x) {
  for (int i = 0; i < p.size(); ++i) {
    int cc = 0;
    int &pp = p[i];
    while (x % pp == 0) {
      x /= pp;
      ++cc;
    }
    dd[i] = max(dd[i], cc);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  if (k == 1) {
    cout << "Yes" << '\n';
    return 0;
  }
  factor(k);
  while (n--) {
    int x;
    cin >> x;
    f(x);
  }
  bool g = true;
  for (int i = 0; i < p.size(); ++i) {
    if (cnt[i] > dd[i]) {
      g = false;
      break;
    }
  }
  cout << (g ? "Yes" : "No") << '\n';
  return 0;
}