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.Arrays;
import java.util.Scanner;



public class Main {

    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[] num=new int[200005];
        int[] cnt=new int[200005];
        for (int i = 0; i <n ; i++) {
            arr[i]=sc.nextInt();
        }
        Arrays.sort(arr);//排序，不然给出的如果不按顺序，循环会出问题
        int ans=Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            int temp=0;
            int x=arr[i];
            while (x!=0)
            {
                //arr[i]出现次数
                cnt[x]++;
                num[x]+=temp;//变成x位置的arr[i]需要操作几步
                if(cnt[x]==k)//当x这个位置==k了说明有符合条件的了，操作一定步数之后，符合题意，步数就是num[x]
                {
                    ans=Math.min(ans,num[x]);
                }
                temp++;
                x/=2;
            }
        }
        System.out.println(ans);

    }
    
}