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 Main {
    public static void main(String args[]) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        in.nextToken();
        int n = (int) in.nval;
        in.nextToken();
        int k = (int) in.nval;
        int a[] = new int[60];
        Vector<Integer> b=new Vector<>();
        for (int i = 1; i <= n; i++) {
            in.nextToken();
            a[i] = (int) in.nval;
            int x=a[i];
            while (x != 0) {
                b.add(x);
                x /= 2;

            }
        }
        int ans = (int)1e9;
        for (int j = 0; j < b.size(); j++) {
            Vector<Integer> c = new Vector<>();
            for (int i = 1; i <= n; i++) {
                int x = a[i], cnt = 0;
                while (x > b.get(j)) {
                    x /= 2;
                    cnt++;
                }
                if (x == b.get(j)) {
                    c.add(cnt);
                }
            }
            if (c.size() < k) continue;
            Collections.sort(c, Comparator.naturalOrder());
            int sum=0;
            for(int i=0;i<k;i++){
                sum+=c.get(i);
            }
            ans = Math.min(ans,sum );
        }
        out.printf("%d\n", ans);
        out.flush();
    }
}