#include <bits/stdc++.h>
using namespace std;
int n;
long long k, t;
vector<long long> nums;
set<long long> marked;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> t;
    nums.push_back(t);
  }
  sort(nums.begin(), nums.end());
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    long long num = nums[i];
    if (marked.count(num) > 0) continue;
    long long kMult = k * num;
    if (binary_search(nums.begin(), nums.end(), kMult)) {
      marked.insert(kMult);
    }
    cnt++;
  }
  cout << cnt << '\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 n;
long long k, t;
vector<long long> nums;
set<long long> marked;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> t;
    nums.push_back(t);
  }
  sort(nums.begin(), nums.end());
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    long long num = nums[i];
    if (marked.count(num) > 0) continue;
    long long kMult = k * num;
    if (binary_search(nums.begin(), nums.end(), kMult)) {
      marked.insert(kMult);
    }
    cnt++;
  }
  cout << cnt << '\n';
}

```