#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
bool valid(int x, int y) { return x >= 0 && y >= 0 && x < 3 && y < 3 ? 1 : 0; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  long long arr[n];
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  int ans = 0;
  bool vis[n + 5];
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      ++ans;
      vis[i] = 1;
      if (binary_search(arr, arr + n, arr[i] * k)) {
        int x = lower_bound(arr, arr + n, arr[i] * k) - arr;
        vis[x] = 1;
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
bool valid(int x, int y) { return x >= 0 && y >= 0 && x < 3 && y < 3 ? 1 : 0; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  long long arr[n];
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  int ans = 0;
  bool vis[n + 5];
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      ++ans;
      vis[i] = 1;
      if (binary_search(arr, arr + n, arr[i] * k)) {
        int x = lower_bound(arr, arr + n, arr[i] * k) - arr;
        vis[x] = 1;
      }
    }
  }
  cout << ans;
  return 0;
}

```