#include <bits/stdc++.h>
using namespace std;
int a[100010], b[100010] = {0}, n, k;
int in(int n, int x) {
  if (x < b[0] || x > b[n - 1]) return false;
  int l = 0, r = n - 1, mid;
  if (x == b[r]) return true;
  while (l < r - 1) {
    mid = (l + r) >> 1;
    if (x < b[mid])
      r = mid;
    else
      l = mid;
  }
  if (x == b[l])
    return true;
  else
    return false;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] % k == 0 && (!in(ans, a[i] / k))) b[ans++] = a[i];
    if (a[i] % k != 0) b[ans++] = a[i];
  }
  cout << ans << endl;
  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;
int a[100010], b[100010] = {0}, n, k;
int in(int n, int x) {
  if (x < b[0] || x > b[n - 1]) return false;
  int l = 0, r = n - 1, mid;
  if (x == b[r]) return true;
  while (l < r - 1) {
    mid = (l + r) >> 1;
    if (x < b[mid])
      r = mid;
    else
      l = mid;
  }
  if (x == b[l])
    return true;
  else
    return false;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] % k == 0 && (!in(ans, a[i] / k))) b[ans++] = a[i];
    if (a[i] % k != 0) b[ans++] = a[i];
  }
  cout << ans << endl;
  return 0;
}

```