#include <bits/stdc++.h>
using namespace std;
bool myCmp(pair<int, int> &p1, pair<int, int> &p2) {
  return p1.first > p2.first;
}
unsigned long long choose(unsigned long long n, unsigned long long k) {
  if (k > n) {
    return 0;
  }
  unsigned long long r = 1;
  for (unsigned long long d = 1; d <= k; ++d) {
    r *= n--;
    r /= d;
  }
  return r;
}
const int MAXN = 10e5;
int arr[MAXN] = {0};
set<int> soso;
long long ans = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    soso.insert(arr[i]);
  }
  sort(arr, arr + n, greater<int>());
  if (n == 1) {
    cout << "1\n";
    return 0;
  }
  if (k == 1) {
    cout << n << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (soso.count(arr[i]) && arr[i] % k == 0 && soso.count(arr[i] / k))
      soso.erase(arr[i] / k);
  }
  cout << soso.size() << endl;
  return 0;
}

### 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;
bool myCmp(pair<int, int> &p1, pair<int, int> &p2) {
  return p1.first > p2.first;
}
unsigned long long choose(unsigned long long n, unsigned long long k) {
  if (k > n) {
    return 0;
  }
  unsigned long long r = 1;
  for (unsigned long long d = 1; d <= k; ++d) {
    r *= n--;
    r /= d;
  }
  return r;
}
const int MAXN = 10e5;
int arr[MAXN] = {0};
set<int> soso;
long long ans = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    soso.insert(arr[i]);
  }
  sort(arr, arr + n, greater<int>());
  if (n == 1) {
    cout << "1\n";
    return 0;
  }
  if (k == 1) {
    cout << n << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (soso.count(arr[i]) && arr[i] % k == 0 && soso.count(arr[i] / k))
      soso.erase(arr[i] / k);
  }
  cout << soso.size() << endl;
  return 0;
}

```