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
#include <bits/stdc++.h>
using namespace std;
int n, k;
int nums[55];
int divs(int num, int target) {
  if (num < target) return 0x3f3f3f3f;
  if (num == target) return 0;
  return 1 + divs(num / 2, target);
}
int main() {
  cin >> n >> k;
  int max_num = 0;
  for (int i = 0; i < n; i++) {
    cin >> nums[i];
    max_num = max(max_num, nums[i]);
  }
  int ans = 0x3f3f3f3f;
  for (int j = 0; j <= max_num; j++) {
    vector<int> qtd_div;
    for (int i = 0; i < n; i++) {
      int log_2 = divs(nums[i], j);
      qtd_div.push_back(log_2);
    }
    sort(qtd_div.begin(), qtd_div.end());
    int swap_min = 0;
    bool is_valid = true;
    for (int i = 0; i < k && is_valid; i++) {
      is_valid = qtd_div[i] < 212345;
      if (!is_valid) break;
      swap_min += qtd_div[i];
    }
    if (!is_valid) continue;
    ans = min(ans, swap_min);
  }
  cout << ans << '\n';
  return 0;
}
