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;
vector<pair<int, int> > V;
int main() {
  int n, k, i, j;
  scanf("%d %d", &n, &k);
  int arr[n];
  long long int suffix_total = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    suffix_total = suffix_total + arr[i];
  }
  sort(arr, arr + n);
  pair<int, int> p;
  p.first = arr[0];
  p.second = 1;
  V.push_back(p);
  for (i = 1; i < n; i++) {
    if (arr[i] == V[V.size() - 1].first) {
      V[V.size() - 1].second++;
    } else {
      p.first = arr[i];
      V.push_back(p);
    }
  }
  if (V[0].second >= k || V[V.size() - 1].second >= k) {
    printf("0");
    return 0;
  }
  long long int ans = 1000000000000000000, prefix_total = 0;
  int prefix_num = 0, suffix_num = n;
  for (i = 0; i < V.size(); i++) {
    if (V[i].second >= k) {
      printf("0");
      return 0;
    }
    int current_num = V[i].second;
    long long int current_sum = V[i].first;
    current_sum = current_sum * current_num;
    suffix_total = suffix_total - current_sum;
    suffix_num = suffix_num - current_num;
    long long int cc = (k - V[i].second);
    long long int first = V[i].first - 1, last = V[i].first + 1;
    first = first * prefix_num - prefix_total;
    last = suffix_total - last * suffix_num;
    if (V[i].second + suffix_num >= k) {
      ans = min(ans, last + cc);
    }
    if (V[i].second + prefix_num >= k) {
      ans = min(ans, first + cc);
    }
    ans = min(ans, first + last + cc);
    prefix_num = prefix_num + current_num;
    prefix_total = prefix_total + current_sum;
  }
  printf("%I64d", ans);
  return 0;
}