#include <bits/stdc++.h>
using namespace std;
void testcases() {}
int bb(long long v[], int n, long long target) {
  int l = 0;
  int r = n - 1;
  int m;
  while (l <= r) {
    m = l + (r - l) / 2;
    if (v[m] == target)
      return m;
    else if (target < v[m])
      r = m - 1;
    else
      l = m + 1;
  }
  return -1;
}
void solve() {
  int n;
  cin >> n;
  int k;
  cin >> k;
  long long v[n];
  long long subset[n];
  int index = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort(v, v + n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (v[i] % k == 0) {
      int pos = bb(subset, index, v[i] / k);
      if (pos == -1) {
        subset[index++] = v[i];
        ans++;
      }
    } else {
      subset[index++] = v[i];
      ans++;
    }
  }
  cout << ans << endl;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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;
void testcases() {}
int bb(long long v[], int n, long long target) {
  int l = 0;
  int r = n - 1;
  int m;
  while (l <= r) {
    m = l + (r - l) / 2;
    if (v[m] == target)
      return m;
    else if (target < v[m])
      r = m - 1;
    else
      l = m + 1;
  }
  return -1;
}
void solve() {
  int n;
  cin >> n;
  int k;
  cin >> k;
  long long v[n];
  long long subset[n];
  int index = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort(v, v + n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (v[i] % k == 0) {
      int pos = bb(subset, index, v[i] / k);
      if (pos == -1) {
        subset[index++] = v[i];
        ans++;
      }
    } else {
      subset[index++] = v[i];
      ans++;
    }
  }
  cout << ans << endl;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}

```