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}.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, dp[N][2];
long long k, a[N];
bool used[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  if (k == 1) {
    cout << n;
    return 0;
  }
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (used[i] == true) {
      continue;
    }
    long long key = a[i];
    vector<int> v;
    v.push_back(0);
    while (true) {
      int left = i - 1, right = n;
      while (right - left > 1) {
        int mid = (left + right) / 2;
        if (a[mid] >= key) {
          right = mid;
        } else {
          left = mid;
        }
      }
      int beg = right;
      left = i, right = n + 1;
      while (right - left > 1) {
        int mid = (left + right) / 2;
        if (a[mid] > key) {
          right = mid;
        } else {
          left = mid;
        }
      }
      if (a[left] == key) {
        v.push_back(left - beg + 1);
        for (int j = beg; j <= left; j++) {
          used[j] = true;
        }
      } else {
        break;
      }
      key = key * k;
    }
    for (int i = 1; i < v.size(); i++) {
      dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]);
      dp[i][1] = dp[i - 1][0] + v[i];
    }
    ans = ans + max(dp[v.size() - 1][0], dp[v.size() - 1][1]);
    for (int i = 1; i < v.size(); i++) {
      dp[i][0] = dp[i][1] = 0;
    }
  }
  cout << ans;
}
