#include <bits/stdc++.h>
using namespace std;
long long exp(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  if (n % 2) {
    long long s = exp(a, n - 1);
    return s * a;
  } else {
    long long s = exp(a, n / 2);
    return s * s;
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<int> arr(n);
  int ans = 0;
  map<int, int> c;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr.begin(), arr.end());
  for (int i = 0; i < n; i++) {
    if (arr[i] % k == 0) {
      if (c[arr[i] / k] == 0) {
        c[arr[i]] = 1;
        ans++;
      }
    } else {
      c[arr[i]] = 1;
      ans++;
    }
  }
  cout << ans << "\n";
}
int main() {
  solve();
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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;
long long exp(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  if (n % 2) {
    long long s = exp(a, n - 1);
    return s * a;
  } else {
    long long s = exp(a, n / 2);
    return s * s;
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<int> arr(n);
  int ans = 0;
  map<int, int> c;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr.begin(), arr.end());
  for (int i = 0; i < n; i++) {
    if (arr[i] % k == 0) {
      if (c[arr[i] / k] == 0) {
        c[arr[i]] = 1;
        ans++;
      }
    } else {
      c[arr[i]] = 1;
      ans++;
    }
  }
  cout << ans << "\n";
}
int main() {
  solve();
  return 0;
}

```