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, k, i, j, m, a[200008], c[200008];
int main() {
  for (scanf("%I64d%I64d", &n, &k); i++ < n; scanf("%I64d", a + i))
    ;
  for (sort(a + 1, a + n + 1), i = 0; i++ < n; c[i] = c[i - 1] + a[i])
    ;
  for (i = n - k; i++ < n; m += a[i] - a[n + 1 - k])
    ;
  for (i = n + 1 - k; a[n + 1 - k] == a[--i]; --m)
    ;
  for (i = 0; i++ < k; j += a[k] - a[i])
    ;
  for (i = k; a[k] == a[++i]; --j)
    ;
  for (j < m ? m = j : 0, i = 1; i <= n && 0 < m;
       j < i + k ? m = min(m, c[n] - (c[i] << 1) + a[i] * (i + i - n) + k - n)
                 : m = 0,
               i = j)
    for (j = i; a[i] == a[++j];)
      ;
  printf("%I64d\n", m < 0 ? 0 : m), exit(0);
}