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 nums[N];
pair<int, int> ggg[N];
int arr[40];
int dp[40][2];
int getDP() {
  for (int i = 0; i < 40; i++) dp[i][0] = dp[i][1] = 0;
  dp[0][1] = arr[0];
  dp[0][0] = 0;
  dp[1][1] = arr[1];
  dp[0][0] = dp[0][1];
  for (int i = 2; i < 40; i++) {
    dp[i][1] = max({dp[i - 1][0], dp[i - 2][0], dp[i - 2][1]}) + arr[i];
    dp[i][0] = max({dp[i - 1][1], dp[i - 1][0], dp[i - 2][0], dp[i - 2][1]});
  }
  for (int i = 0; i < 40; i++) arr[i] = 0;
  return max(dp[39][0], dp[39][1]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  if (k == 1) {
    cout << n << "\n";
    return 0;
  }
  for (int i = 1; i <= n; i++) cin >> nums[i];
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    while (nums[i] % k == 0) {
      nums[i] /= k;
      cnt++;
    }
    ggg[i].first = nums[i];
    ggg[i].second = cnt;
  }
  sort(ggg + 1, ggg + n + 1);
  int prev = -1;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (ggg[i].first == prev) {
      arr[ggg[i].second]++;
    } else {
      ans += getDP();
      prev = ggg[i].first;
      i--;
    }
  }
  ans += getDP();
  cout << ans << "\n";
  return 0;
}
