#include <bits/stdc++.h>
using namespace std;
int flag[100000];
int main() {
  int n, k, i;
  memset(flag, 0, sizeof(flag));
  scanf("%d%d", &n, &k);
  long long a[100000];
  for (i = 0; i < n; i++) scanf("%I64d", &a[i]);
  sort(a, a + n);
  int sum = n;
  for (i = 0; i < n; i++) {
    if (!flag[i]) {
      long long v = k * a[i];
      if (v > a[n - 1]) continue;
      int r, l, mid;
      r = n - 1;
      l = i + 1;
      while (r >= l) {
        mid = (r + l) / 2;
        if (v > a[mid])
          l = mid + 1;
        else if (v < a[mid])
          r = mid - 1;
        else {
          flag[mid] = 1;
          sum--;
          break;
        }
      }
    }
  }
  printf("%d\n", sum);
}

### 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;
int flag[100000];
int main() {
  int n, k, i;
  memset(flag, 0, sizeof(flag));
  scanf("%d%d", &n, &k);
  long long a[100000];
  for (i = 0; i < n; i++) scanf("%I64d", &a[i]);
  sort(a, a + n);
  int sum = n;
  for (i = 0; i < n; i++) {
    if (!flag[i]) {
      long long v = k * a[i];
      if (v > a[n - 1]) continue;
      int r, l, mid;
      r = n - 1;
      l = i + 1;
      while (r >= l) {
        mid = (r + l) / 2;
        if (v > a[mid])
          l = mid + 1;
        else if (v < a[mid])
          r = mid - 1;
        else {
          flag[mid] = 1;
          sum--;
          break;
        }
      }
    }
  }
  printf("%d\n", sum);
}

```