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>
const int INF = (int)(1e9);
const long long INFLL = (long long)(1e18);
const double EPS = 1e-13;
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  map<long long, int> g;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    g[x] = 0;
    if (k == 1) continue;
    if (x % k == 0 && g.count(x / k)) {
      g[x]++;
      g[x / k]++;
    }
    if (g.count(x * k)) {
      g[x]++;
      g[x * k]++;
    }
  }
  int ans = n;
  for (typeof((g).begin()) v = (g).begin(); v != (g).end(); ++v) {
    long long x = v->first;
    if (v->second == 2) {
      if (x % k == 0 && g.count(x / k)) g[x / k]--;
      if (g.count(x * k)) g[x * k]--;
      ans--;
    }
  }
  for (typeof((g).begin()) v = (g).begin(); v != (g).end(); ++v) {
    long long x = v->first;
    if (v->second == 1) {
      if (x % k == 0 && g.count(x / k)) g[x / k]--;
      if (g.count(x * k)) g[x * k]--;
      ans--;
    }
  }
  cout << ans << endl;
}
