#include <bits/stdc++.h>
using namespace std;
const long long maxn = 100010;
long long lis[maxn];
long long l, r, mid, ans;
long long k, n;
bool vis[maxn];
void bs(long long x) {
  while (l <= r) {
    mid = (l + r) / 2;
    long long findd = lis[x] * k;
    if (findd > 1000000000) return;
    if (lis[mid] > findd) {
      r = mid - 1;
    } else if (lis[mid] < findd)
      l = mid + 1;
    else if (lis[mid] == findd) {
      vis[mid] = 1;
      ans--;
      break;
    }
  }
}
int main() {
  scanf("%lld%lld", &n, &k);
  ans = n;
  for (int i = 0; i < n; ++i) scanf("%lld", &lis[i]);
  sort(lis, lis + n);
  if (k != 1) {
    for (int i = 0; i < n; ++i) {
      if (!vis[i]) {
        l = i, r = n;
        bs(i);
      }
    }
  }
  printf("%lld\n", ans);
  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;
const long long maxn = 100010;
long long lis[maxn];
long long l, r, mid, ans;
long long k, n;
bool vis[maxn];
void bs(long long x) {
  while (l <= r) {
    mid = (l + r) / 2;
    long long findd = lis[x] * k;
    if (findd > 1000000000) return;
    if (lis[mid] > findd) {
      r = mid - 1;
    } else if (lis[mid] < findd)
      l = mid + 1;
    else if (lis[mid] == findd) {
      vis[mid] = 1;
      ans--;
      break;
    }
  }
}
int main() {
  scanf("%lld%lld", &n, &k);
  ans = n;
  for (int i = 0; i < n; ++i) scanf("%lld", &lis[i]);
  sort(lis, lis + n);
  if (k != 1) {
    for (int i = 0; i < n; ++i) {
      if (!vis[i]) {
        l = i, r = n;
        bs(i);
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```