#include <bits/stdc++.h>
using namespace std;
const int M = 100010;
set<int> s;
set<int>::iterator it;
int main() {
  int n, k;
  while (cin >> n >> k) {
    s.clear();
    for (int i = 0; i < n; i++) {
      int a;
      scanf("%d", &a);
      s.insert(a);
    }
    if (k == 1) {
      printf("%d\n", n);
      continue;
    }
    int res = 0;
    for (it = s.begin(); it != s.end(); it++) {
      long long m = *it;
      int cnt = 1;
      while (1) {
        m *= k;
        if (m > 1000000000) break;
        if (s.find(m) != s.end()) {
          s.erase(m);
          cnt++;
        } else
          break;
      }
      res += ceil((double)cnt / 2.0);
    }
    cout << res << endl;
  }
}

### Prompt

In Cpp, your task is to solve 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;
const int M = 100010;
set<int> s;
set<int>::iterator it;
int main() {
  int n, k;
  while (cin >> n >> k) {
    s.clear();
    for (int i = 0; i < n; i++) {
      int a;
      scanf("%d", &a);
      s.insert(a);
    }
    if (k == 1) {
      printf("%d\n", n);
      continue;
    }
    int res = 0;
    for (it = s.begin(); it != s.end(); it++) {
      long long m = *it;
      int cnt = 1;
      while (1) {
        m *= k;
        if (m > 1000000000) break;
        if (s.find(m) != s.end()) {
          s.erase(m);
          cnt++;
        } else
          break;
      }
      res += ceil((double)cnt / 2.0);
    }
    cout << res << endl;
  }
}

```