#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using vpii = vector<pair<int, int>>;
using vpll = vector<pair<ll, ll>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
int dx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int dy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
void solve() {
  int n, k;
  cin >> n >> k;
  vi a(n);
  for (ll i = 0; i < n; i++) cin >> a[i];
  ;
  sort(a.begin(), a.end());
  set<int> ans;
  for (ll i = 0; i < (n); i++) {
    if (a[i] % k != 0) {
      ans.insert(a[i]);
    } else {
      if (ans.count(a[i] / k) == 0) ans.insert(a[i]);
    }
  }
  cout << ans.size() << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Your task is to create 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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using vpii = vector<pair<int, int>>;
using vpll = vector<pair<ll, ll>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
int dx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int dy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
void solve() {
  int n, k;
  cin >> n >> k;
  vi a(n);
  for (ll i = 0; i < n; i++) cin >> a[i];
  ;
  sort(a.begin(), a.end());
  set<int> ans;
  for (ll i = 0; i < (n); i++) {
    if (a[i] % k != 0) {
      ans.insert(a[i]);
    } else {
      if (ans.count(a[i] / k) == 0) ans.insert(a[i]);
    }
  }
  cout << ans.size() << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  while (t--) {
    solve();
  }
}

```