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 main() {
  long long int n, k, arr[400005], fac = 0, brr[400005] = {0}, min = LONG_MAX,
                                   cost, initial, final, temp_ans, ans;
  scanf("%lld%lld", &n, &k);
  for (long long int i = 0; i < n; ++i) {
    scanf("%lld", &arr[i]);
    brr[arr[i]]++;
  }
  sort(arr, arr + n);
  printf("\n");
  for (long long int j = 0; j <= arr[n - 1]; ++j) {
    fac = 0;
    cost = 0;
    temp_ans = 0;
    ans = 0;
    initial = j;
    final = j;
    while (ans < k && initial <= 2 * 100000 && final <= 200000) {
      if (fac == 0) {
        if (brr[initial] > 0) {
          temp_ans = ans + brr[initial];
          if (temp_ans < k) {
            ans = temp_ans;
            cost += (fac * brr[initial]);
          } else {
            ans = k;
            cost += (fac * (k - ans));
            break;
          }
        }
      }
      if (ans >= k) break;
      fac++;
      initial = initial * 2;
      final = final * 2 + 1;
      if (initial == 0) initial++;
      for (long long int i = initial;
           i <= final && initial <= 4 * 100000 && final <= 400000; ++i) {
        if (brr[i] > 0) {
          temp_ans = ans + brr[i];
          if (temp_ans <= k) {
            ans = temp_ans;
            cost += (fac * brr[i]);
          } else {
            cost += (fac * (k - ans));
            ans = k;
            break;
          }
        }
        if (ans >= k) break;
      }
      if (j == 2)
        if (ans >= k) break;
    }
    if (ans >= k) {
      if (cost <= min) {
        min = cost;
      }
    }
  }
  printf("%lld\n", min);
}