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 maxN = (int)1E6 + 6;
int n, k, m;
bool avail[maxN];
int a[maxN];
set<int> f, g;
long long GCD(long long x, long long y) {
  if (x < y) return GCD(y, x);
  if (y == 0) return x;
  return GCD(y, x % y);
}
long long LCM(long long x, long long y) { return (x / GCD(x, y)) * y; }
int main() {
  scanf("%d %d", &n, &k);
  memset(avail, false, sizeof(avail));
  for (int i = 1; i <= n; ++i) {
    int c;
    scanf("%d", &c);
    avail[c] = true;
  }
  m = 0;
  for (int i = 1; i <= sqrt(k); ++i) {
    if (k % i == 0) {
      bool ok = false;
      for (int j = i; j < maxN; j += i)
        if (avail[j]) {
          ok = true;
          break;
        }
      if (ok) a[++m] = i;
      ok = false;
      for (int j = k / i; j < maxN; j += (k / i))
        if (avail[j]) {
          ok = true;
          break;
        }
      if (ok) a[++m] = k / i;
    }
  }
  f.clear();
  f.insert(1);
  for (int i = 2; i <= m; ++i) {
    g.clear();
    long long j = a[i];
    for (set<int>::iterator it = f.begin(); it != f.end(); ++it) {
      if (LCM(j, (*it)) <= k && k % LCM(j, (*it)) == 0 &&
          f.find(j * (*it)) == f.end() && g.find(j * (*it)) == g.end())
        g.insert(LCM(j, (*it)));
    }
    for (set<int>::iterator it = g.begin(); it != g.end(); ++it) f.insert(*it);
  }
  if (f.find(k) != f.end())
    puts("Yes");
  else
    puts("No");
  return 0;
}