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;
const int INF = 1e9;
const int MX = 1e6 + 5;
long long MD = 1e9 + 7;
const int MDL = 99824453;
auto time0 = chrono::high_resolution_clock::now();
long long a[MX], LPF[MX];
long long bin_power(long long a, long long b, long long md) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % md;
    a = (a * a) % md;
    b >>= 1;
  }
  return res;
}
long long lcm(long long n) {
  map<long long, long long> mp;
  for (long long i = 1; i <= n; i++) {
    long long cur = a[i], curf = LPF[cur], cnt = 0;
    while (cur > 1) {
      cur /= LPF[cur];
      cnt++;
      if (LPF[cur] != curf) {
        mp[curf] = max(mp[curf], cnt);
        curf = LPF[cur];
        cnt = 0;
      }
    }
  }
  long long ans = 1;
  for (auto x : mp) ans = (ans * bin_power(x.first, x.second, MD)) % MD;
  return ans;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (k == 1) {
    cout << "Yes";
    return;
  }
  bool flag = false;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  MD = k;
  long long L = lcm(n);
  if (L != 0)
    cout << "No";
  else
    cout << "Yes";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  srand(time(0));
  time0 = chrono::high_resolution_clock::now();
  for (long long i = 1; i <= MX - 1; i++) LPF[i] = i;
  for (long long i = 2; i <= MX - 1; i++) {
    if (LPF[i] == i) {
      for (int j = 2 * i; j < MX; j += i)
        if (LPF[j] == j) LPF[j] = i;
    }
  }
  solve();
  cerr << "Execution Time: "
       << chrono::duration_cast<chrono::nanoseconds>(
              chrono::high_resolution_clock::now() - time0)
                  .count() *
              1e-9
       << " sec\n";
  return 0;
}