Problem: You are given the array a consisting of n elements and the integer k ≤ n.

You want to obtain at least k equal elements in the array a. In one move, you can make one of the following two operations:

  * Take one of the minimum elements of the array and increase its value by one (more formally, if the minimum value of a is mn then you choose such index i that a_i = mn and set a_i := a_i + 1); 
  * take one of the maximum elements of the array and decrease its value by one (more formally, if the maximum value of a is mx then you choose such index i that a_i = mx and set a_i := a_i - 1). 



Your task is to calculate the minimum number of moves required to obtain at least k equal elements in the array.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a and the required number of equal elements.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the minimum number of moves required to obtain at least k equal elements in the array.

Examples

Input


6 5
1 2 2 4 2 3


Output


3


Input


7 5
3 3 2 1 1 1 3


Output


4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, k;
int a[200001];
int vis[200001];
int op[200001];
void print(int a[]) {
  for (int i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
}
int main(int argc, char *argv[]) {
  int res = 0x3f3f3f3f;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    vis[a[i]]++;
    if (vis[a[i]] >= k) {
      printf("0\n");
      return 0;
    }
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    int div = 0;
    while (a[i] != 0) {
      a[i] /= 2;
      div++;
      vis[a[i]]++;
      op[a[i]] += div;
      if (vis[a[i]] == k) {
        res = min(res, op[a[i]]);
      }
    }
  }
  printf("%d\n", res);
  return 0;
}