#include <bits/stdc++.h>
using namespace std;
struct s {
  int a1, a2;
} a[100000];
bool comp(s a, s b) {
  if (a.a1 == b.a1)
    return a.a2 < b.a2;
  else
    return a.a1 < b.a1;
}
int main() {
  int n, k, t1, t2, i, z = 1;
  scanf("%d%d", &n, &k);
  if (k == 1) {
    printf("%d\n", n);
    exit(0);
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &t1);
    for (t2 = 0; t1 % k == 0; t1 /= k, t2++)
      ;
    a[i].a1 = t1;
    a[i].a2 = t2;
  }
  sort(a, a + n, comp);
  for (i = 1; i < n; i++) {
    if (a[i].a1 == a[i - 1].a1 && a[i].a2 == a[i - 1].a2 + 1)
      a[i].a1 = 0;
    else
      z++;
  }
  printf("%d\n", z);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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;
struct s {
  int a1, a2;
} a[100000];
bool comp(s a, s b) {
  if (a.a1 == b.a1)
    return a.a2 < b.a2;
  else
    return a.a1 < b.a1;
}
int main() {
  int n, k, t1, t2, i, z = 1;
  scanf("%d%d", &n, &k);
  if (k == 1) {
    printf("%d\n", n);
    exit(0);
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &t1);
    for (t2 = 0; t1 % k == 0; t1 /= k, t2++)
      ;
    a[i].a1 = t1;
    a[i].a2 = t2;
  }
  sort(a, a + n, comp);
  for (i = 1; i < n; i++) {
    if (a[i].a1 == a[i - 1].a1 && a[i].a2 == a[i - 1].a2 + 1)
      a[i].a1 = 0;
    else
      z++;
  }
  printf("%d\n", z);
  return 0;
}

```