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;
const int maxn = 1e6 + 10;
int n, k, cnt;
long long ans, l[maxn], r[maxn], v[maxn], sum[maxn], a[maxn], b[maxn];
int main() {
  scanf("%d%d", &cnt, &k);
  ans = 1e18;
  for (int i = 1; i <= cnt; i++) scanf("%lld", &b[i]);
  sort(b + 1, b + cnt + 1);
  for (int i = 1; i <= cnt; i++) {
    a[++n] = b[i];
    int j = i;
    while (j + 1 <= cnt && b[j + 1] == b[i]) j++;
    v[n] = j - i + 1;
    i = j;
  }
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + v[i];
  l[1] = 0;
  for (int i = 2; i <= n; i++) l[i] = l[i - 1] + sum[i - 1] * (a[i] - a[i - 1]);
  r[n] = 0;
  for (int i = n - 1; i >= 1; i--)
    r[i] = r[i + 1] + (sum[n] - sum[i]) * (a[i + 1] - a[i]);
  for (int i = 1; i <= n; i++)
    if (v[i] >= k)
      ans = min(ans, 0ll);
    else {
      int now = k - v[i];
      if (sum[i - 1] >= now) ans = min(ans, l[i] - (sum[i - 1] - now));
      if (sum[n] - sum[i] >= now)
        ans = min(ans, r[i] - (sum[n] - sum[i] - now));
      ans = min(ans, l[i] + r[i] - (sum[n] - v[i] - now));
    }
  printf("%lld\n", ans);
  return 0;
}