#include <bits/stdc++.h>
using namespace std;
int n, k, a[100001];
long long b[100001];
bool has[100001];
bool can(int mid) {
  memset(has, false, sizeof has);
  int tm = 0;
  for (int i = 0; i < (int)(n); ++i) {
    if (!has[i]) {
      tm++;
      int t = lower_bound(a, a + n, b[i]) - a;
      if (t < n && a[t] == b[i]) has[t] = true;
    }
    if (tm >= mid) return true;
  }
  return tm >= mid;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < (int)(n); ++i) {
    scanf("%d", a + i);
    b[i] = 1ll * a[i] * k;
  }
  sort(a, a + n);
  sort(b, b + n);
  int l = 0, r = n;
  long long res = 0;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (can(mid))
      res = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%lld\n", res);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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, a[100001];
long long b[100001];
bool has[100001];
bool can(int mid) {
  memset(has, false, sizeof has);
  int tm = 0;
  for (int i = 0; i < (int)(n); ++i) {
    if (!has[i]) {
      tm++;
      int t = lower_bound(a, a + n, b[i]) - a;
      if (t < n && a[t] == b[i]) has[t] = true;
    }
    if (tm >= mid) return true;
  }
  return tm >= mid;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < (int)(n); ++i) {
    scanf("%d", a + i);
    b[i] = 1ll * a[i] * k;
  }
  sort(a, a + n);
  sort(b, b + n);
  int l = 0, r = n;
  long long res = 0;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (can(mid))
      res = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%lld\n", res);
  return 0;
}

```