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 comment(linker, "/STACK:10240000,10240000")
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-10;
inline int read() {
  static char ch;
  bool sgn = false;
  while (ch = getchar(), ch < '0' || ch > '9')
    if (ch == '-') sgn = true;
  int res = ch - 48;
  while (ch = getchar(), ch >= '0' && ch <= '9') res = res * 10 + ch - 48;
  return sgn ? -res : res;
}
const int N = 1e6 + 5;
int a[N], t;
bool bo[N];
int p[N];
vector<pair<int, int> > pr;
int main() {
  int n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  memset(bo, 1, sizeof(bo));
  for (int i = 2; i <= 1000000; i++) {
    if (bo[i]) {
      t++;
      p[t] = i;
    }
    for (int j = 1; (j <= t) && (i * p[j] <= 1000000); j++) {
      bo[i * p[j]] = 0;
      if (i % p[j] == 0) break;
    }
  }
  for (int i = 1; i <= t; i++) {
    if (k % p[i] == 0) {
      int s = 0;
      while (k % p[i] == 0) {
        s++;
        k /= p[i];
      }
      pr.push_back(make_pair(p[i], s));
    }
  }
  int s = pr.size();
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < s; j++) {
      if (pr[j].second == 0) continue;
      int t = 0, p = pr[j].first;
      while (a[i] % p == 0) {
        t++;
        a[i] /= p;
      }
      if (t >= pr[j].second) pr[j].second = 0;
    }
  }
  for (int i = 0; i < s; i++)
    if (pr[i].second > 0) return puts("No");
  puts("Yes");
}