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;
using namespace std;
void enumerateSubmasks(long long m) {
  for (long long s = m;; s = (s - 1) & m) {
    if (s == 0) {
      break;
    }
  }
}
vector<long long> primes;
long long sz = 1e6 + 1;
vector<bool> val(1000001, true);
vector<long long> fP(1000001, 0);
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  val[0] = 0;
  val[1] = 0;
  for (long long i = 2; i < sz; i++) {
    if (!val[i]) continue;
    fP[i] = i;
    primes.push_back(i);
    for (long long j = i * i; j < sz; j += i) {
      if (val[j]) fP[j] = i;
      val[j] = false;
    }
  }
  vector<long long> cnt(1000001, 0);
  long long n, k;
  cin >> n >> k;
  vector<long long> arr(n);
  for (long long i = 0; i < n; ++i) cin >> arr[i];
  for (long long i = 0; i < n; ++i) {
    long long temp = arr[i];
    long long currCount = 0;
    long long currPrime = fP[temp];
    while (temp != 1) {
      if (fP[temp] == currPrime) {
        currCount += 1;
        temp /= currPrime;
      } else {
        cnt[currPrime] = max(cnt[currPrime], currCount);
        currCount = 1;
        currPrime = fP[temp];
        temp /= currPrime;
      }
    }
    cnt[currPrime] = max(cnt[currPrime], currCount);
  }
  for (auto node : primes) {
    long long capacity = cnt[node];
    while ((k % node == 0) && capacity) {
      k /= node;
      capacity--;
    }
  }
  if (k == 1) {
    cout << "Yes" << endl;
  } else
    cout << "No" << endl;
}