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;
long long n, m, a[200005], b[200005];
int main() {
  long long int i, j, c = 0, t, l, r, mid, z, k;
  scanf("%lld", &n);
  scanf("%lld", &m);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  sort(a + 1, a + n + 1);
  for (i = 1; i <= n; i++) {
    b[i - 1] = a[i];
    a[i] += a[i - 1];
  }
  long long ans = 1e18;
  for (i = 1; i <= n; i++) {
    c = a[i] - a[i - 1];
    long long same = upper_bound(b, b + n, c) - lower_bound(b, b + n, c);
    long long chuto = upper_bound(b, b + n, c - 1) - b;
    long long boro = n - (upper_bound(b, b + n, c) - b);
    if (same >= m) {
      printf("0");
      return 0;
    }
    k = m - same;
    long long chutoMove = chuto * (c - 1) - a[chuto];
    long long boroMove = a[n] - a[n - boro] - boro * (c + 1);
    if (chuto >= k) {
      ans = min(ans, chutoMove + k);
    }
    if (boro >= k) {
      ans = min(ans, boroMove + k);
    }
    ans = min(ans, boroMove + chutoMove + k);
  }
  printf("%lld\n", ans);
}