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;
template <typename T, typename U>
inline void chkmax(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline void chkmin(T& x, U y) {
  if (y < x) x = y;
}
bool ip[1000006];
int po[1000006];
int P[1000000], sz, c[1000006];
int n, k;
vector<pair<int, int> > V;
void init() {
  for (int i = 0; i < 1000006; i++) ip[i] = 1;
  ip[0] = ip[1] = 0;
  for (int i = 2; i <= 1002; i++) {
    if (!ip[i]) continue;
    for (int j = i * i; j < 1000006; j += i) ip[j] = 0;
  }
  for (int i = 2; i < 1000006; i++)
    if (ip[i]) P[sz++] = i;
}
int main() {
  init();
  cin >> n >> k;
  for (int i = 0; i < n; i++) scanf("%d", c + i);
  int sk = sqrt(k) + 2;
  for (int i = 2; i < sk; i++) {
    if (ip[i])
      while (k % i == 0) po[i]++, k /= i;
  }
  if (k > 1) po[k]++;
  for (int i = 2; i < 1000006; i++)
    if (po[i]) V.push_back(pair<int, int>(i, po[i]));
  for (int i = 0; i < V.size(); i++) {
    int pp = V[i].first;
    int poo = V[i].second;
    int rlt = 0;
    for (int j = 0; j < n; j++) {
      int v = 0;
      while (c[j] % pp == 0) v++, c[j] /= pp;
      chkmax(rlt, v);
    }
    if (poo > rlt) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
  return 0;
}