#include <bits/stdc++.h>
using namespace std;
long long MOD = 1e9 + 7;
const int OO = (int)1e9 + 7;
const int MAX = 109;
char c1, c2, c3;
string s1, s2, s3;
long long n, m, k, t, a, b, c, d, e, f, x, y, z, sol = 0, ans = 0;
long long ar[1000009];
int dp[1000009][2];
map<long long, bool> vis;
int fun(int i) {
  if (i == n) return 0;
  bool l = 1;
  if (ar[i] % m == 0 && vis[ar[i] / m] == 1) {
    l = 0;
  }
  int &ret = dp[i][l];
  if (ret != -1) return ret;
  ret = 0;
  if (l == 1) {
    vis[ar[i]] = 1;
    ret = max(ret, fun(i + 1) + 1);
    vis[ar[i]] = 0;
  }
  ret = max(ret, fun(i + 1));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  memset((dp), (-1), sizeof(dp));
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> ar[i];
  sort(ar, ar + n);
  cout << fun(0);
  cout << "\n\n";
  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;
long long MOD = 1e9 + 7;
const int OO = (int)1e9 + 7;
const int MAX = 109;
char c1, c2, c3;
string s1, s2, s3;
long long n, m, k, t, a, b, c, d, e, f, x, y, z, sol = 0, ans = 0;
long long ar[1000009];
int dp[1000009][2];
map<long long, bool> vis;
int fun(int i) {
  if (i == n) return 0;
  bool l = 1;
  if (ar[i] % m == 0 && vis[ar[i] / m] == 1) {
    l = 0;
  }
  int &ret = dp[i][l];
  if (ret != -1) return ret;
  ret = 0;
  if (l == 1) {
    vis[ar[i]] = 1;
    ret = max(ret, fun(i + 1) + 1);
    vis[ar[i]] = 0;
  }
  ret = max(ret, fun(i + 1));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  memset((dp), (-1), sizeof(dp));
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> ar[i];
  sort(ar, ar + n);
  cout << fun(0);
  cout << "\n\n";
  return 0;
}

```