#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
vector<int> v(N);
bool vis[N];
long long temp = 0;
void dfs(int s) {
  vis[s] = 1;
  temp++;
  if (v[s]) dfs(v[s]);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  int a[n + 1];
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    long long yo = k * a[i];
    long long l = i + 1;
    long long r = n;
    while (l < r) {
      long long mid = l + ((r - l) >> 1LL);
      if (a[mid] < yo)
        l = mid + 1;
      else
        r = mid;
    }
    if (a[l] == yo) v[i] = l;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (v[i] && !vis[i]) {
      temp = 0;
      dfs(i);
      ans += temp / 2 + temp % 2;
    }
    if (!vis[i]) ans++;
  }
  cout << ans;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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}.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
vector<int> v(N);
bool vis[N];
long long temp = 0;
void dfs(int s) {
  vis[s] = 1;
  temp++;
  if (v[s]) dfs(v[s]);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  int a[n + 1];
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    long long yo = k * a[i];
    long long l = i + 1;
    long long r = n;
    while (l < r) {
      long long mid = l + ((r - l) >> 1LL);
      if (a[mid] < yo)
        l = mid + 1;
      else
        r = mid;
    }
    if (a[l] == yo) v[i] = l;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (v[i] && !vis[i]) {
      temp = 0;
      dfs(i);
      ans += temp / 2 + temp % 2;
    }
    if (!vis[i]) ans++;
  }
  cout << ans;
  return 0;
}

```