#include <bits/stdc++.h>
using namespace std;
int n, k;
long long a[100005];
bool tag[100005];
int main() {
  while (scanf("%d %d", &n, &k) != EOF) {
    for (int i = 0; i < n; i++) scanf("%I64d", a + i);
    memset(tag, false, sizeof(tag));
    sort(a, a + n);
    int ans = n;
    for (int i = 0; i < n; i++) {
      if (!tag[i]) {
        long long p = a[i] * k;
        if (p > a[n - 1]) continue;
        int low = i + 1, high = n - 1, mid;
        while (low <= high) {
          mid = (low + high) / 2;
          if (a[mid] > p)
            high = mid - 1;
          else if (a[mid] < p)
            low = mid + 1;
          else {
            tag[mid] = true;
            ans--;
            break;
          }
        }
      }
    }
    printf("%d\n", ans);
  }
  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;
int n, k;
long long a[100005];
bool tag[100005];
int main() {
  while (scanf("%d %d", &n, &k) != EOF) {
    for (int i = 0; i < n; i++) scanf("%I64d", a + i);
    memset(tag, false, sizeof(tag));
    sort(a, a + n);
    int ans = n;
    for (int i = 0; i < n; i++) {
      if (!tag[i]) {
        long long p = a[i] * k;
        if (p > a[n - 1]) continue;
        int low = i + 1, high = n - 1, mid;
        while (low <= high) {
          mid = (low + high) / 2;
          if (a[mid] > p)
            high = mid - 1;
          else if (a[mid] < p)
            low = mid + 1;
          else {
            tag[mid] = true;
            ans--;
            break;
          }
        }
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

```