#include <bits/stdc++.h>
using namespace std;
const long long int inf = 1e18;
const long long int mod = 1e9 + 7;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long int t, n, k, p, c, d, x, s = 0, i;
  cin >> n >> k;
  vector<long long int> v;
  set<long long int> ins, del;
  for (long long int i = 0; i < n; i++) {
    cin >> p;
    v.push_back(p);
  }
  sort(v.begin(), v.end());
  for (long long int i = 0; i < n; i++) {
    if (del.find(v[i]) == del.end()) {
      ins.insert(v[i]);
      del.insert(v[i] * k);
    }
  }
  cout << ins.size() << endl;
}

### 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;
const long long int inf = 1e18;
const long long int mod = 1e9 + 7;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long int t, n, k, p, c, d, x, s = 0, i;
  cin >> n >> k;
  vector<long long int> v;
  set<long long int> ins, del;
  for (long long int i = 0; i < n; i++) {
    cin >> p;
    v.push_back(p);
  }
  sort(v.begin(), v.end());
  for (long long int i = 0; i < n; i++) {
    if (del.find(v[i]) == del.end()) {
      ins.insert(v[i]);
      del.insert(v[i] * k);
    }
  }
  cout << ins.size() << endl;
}

```