#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (a == 0 || b == 0) return a + b;
  return a / b + gcd(b, a % b);
}
void solve(long long int T) {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort(v.begin(), v.end());
  vector<long long int> ans;
  for (long long int i = 0; i < n; i++) {
    long long int val = v[i];
    if (val % k) {
      ans.emplace_back(val);
    } else {
      if (binary_search(ans.begin(), ans.end(), val / k)) continue;
      ans.emplace_back(val);
    }
  }
  cout << ans.size() << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T = 1;
  for (long long int t = 1; t <= T; t++) {
    solve(t);
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << 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;
long long int gcd(long long int a, long long int b) {
  if (a == 0 || b == 0) return a + b;
  return a / b + gcd(b, a % b);
}
void solve(long long int T) {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort(v.begin(), v.end());
  vector<long long int> ans;
  for (long long int i = 0; i < n; i++) {
    long long int val = v[i];
    if (val % k) {
      ans.emplace_back(val);
    } else {
      if (binary_search(ans.begin(), ans.end(), val / k)) continue;
      ans.emplace_back(val);
    }
  }
  cout << ans.size() << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T = 1;
  for (long long int t = 1; t <= T; t++) {
    solve(t);
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
  return 0;
}

```