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 = 0x3f3f3f3f;
bool vis[1000000 + 10];
long long a[1000000 + 10];
int n;
int ma;
int k;
int ans = 1;
int hehe = 1;
int f[100000 + 10];
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
void work(int tem) {
  int m = sqrt(tem + 0.5);
  for (int i = 2; i <= m && tem > 1; i++) {
    while (tem % i == 0 && tem > 1) {
      f[i]++;
      tem /= i;
      if (f[i] == n) {
        ans *= i;
        f[i] -= n;
      }
    }
  }
  if (tem > 1) {
    f[tem]++;
    if (f[tem] == n) {
      f[tem] -= n;
      ans *= tem;
    }
  }
}
int main() {
  a[0] = 1;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    a[i] = a[i - 1] / gcd(a[i - 1], a[i]) * a[i];
    a[i] %= k;
  }
  hehe = (a[n] + 1) % k;
  if (vis[k]) {
    printf("Yes\n");
    return 0;
  }
  if ((1 % k) != hehe)
    printf("No\n");
  else
    printf("Yes\n");
  return 0;
}