A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.

You're given a set of n distinct positive integers. Your task is to find the size of it's largest k-multiple free subset.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109). The next line contains a list of n distinct positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

All the numbers in the lines are separated by single spaces.

Output

On the only line of the output print the size of the largest k-multiple free subset of {a1, a2, ..., an}.

Examples

Input

6 2
2 3 6 5 4 10


Output

3

Note

In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}.
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y);
long long int gcd(long long int a, long long int b);
vector<long long int> ft = {1};
long long int modInv(long long int i) { return power(i, 1000000007 - 2); }
long long int ncr(long long int n, long long int r) {
  return (n >= r ? (ft[n] * modInv(ft[r])) % 1000000007 * modInv(ft[n - r]) %
                       1000000007
                 : 0);
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  long long int a[n];
  map<long long int, long long int> m1;
  for (long long int i = 0; i < n; i++) cin >> a[i], m1[a[i]]++;
  sort(a, a + n);
  map<long long int, long long int> m;
  long long int ans = 0;
  if (k == 1) {
    cout << n << "\n";
    return 0;
  }
  for (long long int i = 0; i < n; i++) {
    if (a[i] % k == 0) {
      if (m[a[i] / k] == 0 and m1[a[i] / k]) {
        ans++;
        m[a[i]]++;
      }
    }
  }
  cout << n - ans << "\n";
}
