#include <bits/stdc++.h>
using namespace std;
int returnDiv(unordered_map<int, int> &MAP, int x, int k) {
  if (MAP[x] == -1)
    return 0;
  else if (x < k || x % k != 0) {
    MAP[x] = -1;
    return 1;
  } else if (MAP.find(x / k) == MAP.end()) {
    MAP[x] = -1;
    return 1;
  } else {
    MAP[x] = -1;
    return 1 + returnDiv(MAP, x / k, k);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<int> ARR(n);
  unordered_map<int, int> MAP;
  for (int &x : ARR) {
    cin >> x;
    MAP[x] = 1;
  }
  int count = 0;
  sort(ARR.begin(), ARR.end(), [](int a, int b) { return a > b; });
  for (int i = 0; i < n; ++i) {
    int countCurr = returnDiv(MAP, ARR[i], k);
    if (countCurr > 0 && (countCurr & 1)) {
      ++countCurr;
    }
    count += countCurr / 2;
  }
  cout << count << '\n';
}

### Prompt

Your challenge is to write 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;
int returnDiv(unordered_map<int, int> &MAP, int x, int k) {
  if (MAP[x] == -1)
    return 0;
  else if (x < k || x % k != 0) {
    MAP[x] = -1;
    return 1;
  } else if (MAP.find(x / k) == MAP.end()) {
    MAP[x] = -1;
    return 1;
  } else {
    MAP[x] = -1;
    return 1 + returnDiv(MAP, x / k, k);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<int> ARR(n);
  unordered_map<int, int> MAP;
  for (int &x : ARR) {
    cin >> x;
    MAP[x] = 1;
  }
  int count = 0;
  sort(ARR.begin(), ARR.end(), [](int a, int b) { return a > b; });
  for (int i = 0; i < n; ++i) {
    int countCurr = returnDiv(MAP, ARR[i], k);
    if (countCurr > 0 && (countCurr & 1)) {
      ++countCurr;
    }
    count += countCurr / 2;
  }
  cout << count << '\n';
}

```