#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e5 + 5;
long long N, K;
long long v[MAXN];
int main() {
  cin >> N >> K;
  for (int i = 0; i < N; i++) cin >> v[i];
  sort(v, v + N);
  long long resp = N;
  for (int i = 0; i < N; i++) {
    long long value = v[i] * K;
    int l = i + 1, r = N - 1;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (v[mid] == value) {
        resp--;
        v[mid] = -1;
        break;
      }
      if (v[mid] < value)
        l = mid + 1;
      else
        r = mid - 1;
    }
  }
  cout << resp << endl;
  return 0;
}

### Prompt

Generate 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;
const long long MAXN = 1e5 + 5;
long long N, K;
long long v[MAXN];
int main() {
  cin >> N >> K;
  for (int i = 0; i < N; i++) cin >> v[i];
  sort(v, v + N);
  long long resp = N;
  for (int i = 0; i < N; i++) {
    long long value = v[i] * K;
    int l = i + 1, r = N - 1;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (v[mid] == value) {
        resp--;
        v[mid] = -1;
        break;
      }
      if (v[mid] < value)
        l = mid + 1;
      else
        r = mid - 1;
    }
  }
  cout << resp << endl;
  return 0;
}

```