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 = 2e5 + 10;
long long raw[maxn], pre[maxn], suf[maxn], cnt;
long long ans = 1e17;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%lld", &raw[i]);
  sort(raw + 1, raw + n + 1);
  for (int i = 1; i <= n; i++) {
    if (raw[i] != raw[i - 1]) {
      if (cnt >= k) {
        printf("0");
        return 0;
      }
      cnt = 1;
    } else
      cnt++;
  }
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + raw[i];
  for (int i = n; i >= 1; i--) suf[i] = suf[i + 1] + raw[i];
  for (int i = 1; i <= n; i++) {
    if (i < k)
      ans = min(ans,
                i * raw[i] - pre[i] + suf[i] - (n - i + 1) * raw[i] - (n - k));
    else
      ans = min(ans, i * raw[i] - pre[i] - (i - k));
    if (n - i + 1 < k)
      ans = min(ans,
                suf[i] - (n - i + 1) * raw[i] + i * raw[i] - pre[i] - (n - k));
    else
      ans = min(ans, suf[i] - (n - i + 1) * raw[i] - (n - i + 1 - k));
  }
  cout << max(ans, 0ll);
  return 0;
}