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>
int setBit(int N, int pos) { return N = N | (1 << pos); }
int resetBit(int N, int pos) { return N = N & ~(1 << pos); }
bool checkBit(int N, int pos) { return (bool)(N & (1 << pos)); }
using namespace std;
long long n, k, key, arr[100010], dp[100010], TC, flag[100010];
map<long long, long long> mapp;
vector<long long> graph[100010];
bool visit[100010];
long long binSearch(long long val) {
  long long low = 0, high = n - 1, mid;
  while (low <= high) {
    mid = (low + high) / 2;
    if (arr[mid] == val)
      return mid;
    else if (arr[mid] > val)
      high = mid - 1;
    else
      low = mid + 1;
  }
}
long long DP(long long idx, long long x, long long len) {
  if (idx >= len) return 0;
  if (flag[idx] == TC) return dp[idx];
  flag[idx] = TC;
  long long ret1 = 0, ret2 = 0;
  ret1 = graph[x][idx] + DP(idx + 2, x, len);
  ret2 = DP(idx + 1, x, len);
  return dp[idx] = max(ret1, ret2);
}
int main() {
  long long i, j, ret, mx = -1, temp, ans = 0;
  scanf("%I64d %I64d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%I64d", &arr[i]);
    mx = max(mx, arr[i]);
    if (mapp.find(arr[i]) == mapp.end())
      mapp[arr[i]] = 1;
    else
      mapp[arr[i]]++;
  }
  sort(arr, arr + n);
  for (i = 0; i < n; i++) {
    if (mapp[arr[i]] == -1) continue;
    temp = arr[i];
    while (temp <= mx) {
      if (mapp.find(temp) == mapp.end() || mapp[temp] == -1) break;
      graph[key].push_back(mapp[temp]);
      mapp[temp] = -1;
      temp *= k;
    }
    key++;
  }
  for (i = 0; i < key; i++) {
    TC++;
    ret = DP(0, i, (int)graph[i].size());
    ans += ret;
  }
  printf("%I64d\n", ans);
  return 0;
}
