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 JAVA):

Solution:

import java.util.*;

public class codeforces {


    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();

        int[] arr = new int[n];
        int max = 0;
        for(int i=0;i<n;i++) {
            arr[i] = sc.nextInt();
            max = Math.max(max,arr[i]);
        }

        
        ArrayList<Integer>[] matrix = new ArrayList[max+1];

        for(int i=0;i<=max;i++)
            matrix[i] = new ArrayList<>();

        for(int i=0;i<n;i++)
        {
            int temp = arr[i];
            int count = 0;

            while(true)
            {
                matrix[temp].add(count);
                if(temp == 0)
                    break;
                temp = temp / 2;
                count++;
            }
        }



        long ans = Long.MAX_VALUE;
        for(int i=0;i<=max;i++)
        {
            long temp = 0;
            long element = 0;

            if (matrix[i].size()<k) continue;
            Collections.sort(matrix[i]);
            for(int j=0;j<k;j++)
            {
                temp = temp + matrix[i].get(j);
            }
            ans = Math.min(ans,temp);
        }
        System.out.println(ans);


    }



}