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, m, cnt;
long long sum[200100], csum[200100], total, mk[200100], k,
    ans = 999999999999999;
int a[200100], b[200100];
map<int, int> mp;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
    total += a[i];
  }
  sort(1 + b, 1 + b + n);
  for (int i = 1; i <= n; i++) {
    if (mp.find(b[i]) == mp.end()) {
      mp[b[i]] = ++cnt;
      mk[cnt] = b[i];
    }
    csum[cnt] += b[i];
    sum[cnt]++;
  }
  bool flag = false;
  for (int i = 1; i <= cnt; i++) {
    if (sum[i] >= m) flag = true;
    csum[i] += csum[i - 1];
    sum[i] += sum[i - 1];
  }
  if (flag) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i <= cnt; i++) {
    if (sum[i] >= m) {
      k = sum[i] * mk[i] - (sum[i] - m) - csum[i];
      ans = min(ans, k);
    }
    if (n - sum[i - 1] >= m) {
      k = total - csum[i - 1] -
          ((n - sum[i - 1]) * mk[i] + ((n - sum[i - 1]) - m));
      ans = min(ans, k);
    }
    k = (sum[i - 1] * mk[i] - csum[i - 1]) +
        ((total - csum[i]) - (n - sum[i]) * mk[i]) - (n - m);
    ans = min(ans, k);
  }
  printf("%lld\n", ans);
  return 0;
}