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.Collections;
import java.util.PriorityQueue;
import java.util.Scanner;

public class EqualizingByDivision {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		int n=scn.nextInt();
		int k=scn.nextInt();
        PriorityQueue<Integer> []k_windos=new PriorityQueue[200006];
        int []opns=new int[200006];
        int max=0;
        for(int i=0;i<200006;i++) {
        	k_windos[i]=new PriorityQueue<>(Collections.reverseOrder());
        }
        for(int i=0;i<n;i++) {
        	int temp=scn.nextInt();
        	int cnt=0;
        	if(max<temp) {
        		max=temp;
        	}
        	while(temp!=0) {
        		
        		//System.out.println("temp "+temp);

        		if(!k_windos[temp].isEmpty()) {
        			//System.out.println(i+" ch "+cnt);
        			PriorityQueue<Integer> q=k_windos[temp];
        			if(q.size()+1<=k) {
        				q.add(cnt);
        				opns[temp]+=cnt;
        			}else {
        				int x=q.peek();
        				//System.out.println(q);
        				if(cnt<x) {
        					q.poll();
        					q.add(cnt);
        					opns[temp]=opns[temp]-x+cnt;
        				}
        			}
        		}else {
        			PriorityQueue<Integer> q=new PriorityQueue<>(Collections.reverseOrder());
        			q.add(cnt);
        			k_windos[temp]=q;
        			//System.out.println(q+" q "+cnt);
        			opns[temp]=cnt;
        		}
        		cnt++;
        		temp/=2;
        	}
        	
        }
        int min=Integer.MAX_VALUE;
        int flag=0;
        for(int i=0;i<=max;i++) {
        	if(opns[i]<min&&k_windos[i].size()==k) {
        		//System.out.println("kkkk "+i);
        		flag=1;
        		min=opns[i];
        	}
        }
        if(flag==0) {
        	System.out.println(0);
        }else {
        	System.out.println(min);
        }
       // System.out.println(k_windos[2]+" "+k_windos[1]+" "+k_windos[3]+" "+opns[2]+" "+opns[1]+" "+opns[3]+" "+max+" "+k_windos[3].size());
	}

}