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.io.*;
import java.util.*;

public class D
{
    PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok;

    public void go() throws IOException
    {
        ntok();
        int n = ipar();
        int k = ipar();
        ntok();
        int[] arr = new int[n];
        int[] count = new int[500001];
        ArrayList<PriorityQueue<Integer>> div = new ArrayList<>();
        for (int i = 0; i < 500001; i++)
        {
            div.add(new PriorityQueue<>());
        }
        for (int i = 0; i < n; i++)
        {
            arr[i] = ipar();
            int curr = arr[i];
            int d = 0;
            while (curr > 0)
            {
                count[curr]++;
                div.get(curr).add(d);
                curr /= 2;
                d++;
            }
            count[0]++;
            div.get(0).add(d);
        }
        int min = 1000000000;
        for (int i = 0; i < count.length; i++)
        {
            if (count[i] >= k)
            {
                int total = 0;
                for (int e = 0; e < k; e++)
                {
                    total += div.get(i).remove();
                }
                min = Math.min(total, min);
            }
        }
        out.println(min);

        out.flush();
        in.close();
    }

    public void ntok() throws IOException
    {
        tok = new StringTokenizer(in.readLine());
    }

    public int ipar()
    {
        return Integer.parseInt(tok.nextToken());
    }

    public long lpar()
    {
        return Long.parseLong(tok.nextToken());
    }

    public double dpar()
    {
        return Double.parseDouble(tok.nextToken());
    }

    public String spar()
    {
        return tok.nextToken();
    }

    public static void main(String[] args) throws IOException
    {
        new D().go();
    }
}