#include <bits/stdc++.h>
using namespace std;
const int N_MAX = 100000;
long long a[N_MAX];
bool used[N_MAX];
long long n, k, ans;
int binSearch(long long q) {
  int l = 0, r = n, m = l + (r - l) / 2;
  while (l < r) {
    if (q <= a[m]) {
      r = m;
    } else {
      l = m + 1;
    }
    m = (l + r) / 2;
  }
  if (a[r] == q) {
    return r;
  } else {
    return -1;
  }
}
int main() {
  cin >> n >> k;
  ans = n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if (!used[i]) {
      int res = binSearch(a[i] * 1ll * k);
      if (res != -1) {
        ans--;
        used[res] = true;
      }
    }
  }
  cout << max(n - ans, ans);
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const int N_MAX = 100000;
long long a[N_MAX];
bool used[N_MAX];
long long n, k, ans;
int binSearch(long long q) {
  int l = 0, r = n, m = l + (r - l) / 2;
  while (l < r) {
    if (q <= a[m]) {
      r = m;
    } else {
      l = m + 1;
    }
    m = (l + r) / 2;
  }
  if (a[r] == q) {
    return r;
  } else {
    return -1;
  }
}
int main() {
  cin >> n >> k;
  ans = n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if (!used[i]) {
      int res = binSearch(a[i] * 1ll * k);
      if (res != -1) {
        ans--;
        used[res] = true;
      }
    }
  }
  cout << max(n - ans, ans);
  return 0;
}

```