#include <bits/stdc++.h>
void solve(int n) {
  int i, j, l, k, m, res;
  std::vector<std::vector<int>> v;
  std::vector<int> one(2);
  scanf("%d", &m);
  if (m == 1) {
    printf("%d", n);
    return;
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &l);
    k = 0;
    while (l % m == 0) {
      k++;
      l /= m;
    }
    one[0] = l;
    one[1] = k;
    v.push_back(one);
  }
  std::sort(v.begin(), v.end());
  one[0] = -1;
  v.push_back(one);
  res = 0;
  for (i = 0; i < n; i++) {
    res++;
    if (v[i][0] == v[i + 1][0] && v[i][1] + 1 == v[i + 1][1]) i++;
  }
  printf("%d\n", res);
}
int main() {
#pragma comment(linker, "/STACK:268435456")
  int n, m;
  scanf("%d", &n);
  solve(n);
  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>
void solve(int n) {
  int i, j, l, k, m, res;
  std::vector<std::vector<int>> v;
  std::vector<int> one(2);
  scanf("%d", &m);
  if (m == 1) {
    printf("%d", n);
    return;
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &l);
    k = 0;
    while (l % m == 0) {
      k++;
      l /= m;
    }
    one[0] = l;
    one[1] = k;
    v.push_back(one);
  }
  std::sort(v.begin(), v.end());
  one[0] = -1;
  v.push_back(one);
  res = 0;
  for (i = 0; i < n; i++) {
    res++;
    if (v[i][0] == v[i + 1][0] && v[i][1] + 1 == v[i + 1][1]) i++;
  }
  printf("%d\n", res);
}
int main() {
#pragma comment(linker, "/STACK:268435456")
  int n, m;
  scanf("%d", &n);
  solve(n);
  return 0;
}

```