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 long long mod = 1000000007, mod2 = 2e9;
double Pi = 3.141592653589793238;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
unsigned long long FP(unsigned long long b, unsigned long long p) {
  if (p == 0) return 1;
  return (p % 2 == 1) ? (b)*FP(b * b, p / 2) : FP(b * b, p / 2);
}
long long T, K;
long long BS(long long arr[], long long x);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> T >> K;
  long long arr[T], Fans = T, x;
  for (int i = (0); i < (T); i++) {
    cin >> arr[i];
  }
  if (K == 1) {
    cout << Fans;
    return 0;
  }
  sort(arr, arr + T);
  for (int i = (0); i < (T); i++) {
    x = arr[i] * K;
    if (x == 0) continue;
    long long location = BS(arr, x);
    if (location == -1) {
      continue;
    } else {
      arr[location] = 0;
      Fans--;
    }
  }
  cout << Fans;
}
long long BS(long long arr[], long long x) {
  int lo = 0;
  int hi = T - 1;
  long long mid;
  while (lo <= hi) {
    mid = (lo + hi) / 2;
    if (arr[mid] < x) {
      lo = mid + 1;
    } else if (arr[mid] > x) {
      hi = mid - 1;
    } else {
      return mid;
    }
  }
  return -1;
}
