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.lang.*;
import java.util.Scanner;
public class Main{
	public static void main(String[] args) {
		try(Scanner in = new Scanner(System.in)){
			int [] count = new int [200100];
			int [] sort_count = new int [200100];
			int n,k,i,x,temp;
			n=in.nextInt();
			k=in.nextInt();
			int [] num = new int [n]; 
			for(i=0; i<n; i++)
				num[i]=in.nextInt();
			Arrays.sort(num);
			int min=Integer.MAX_VALUE;
			for( i=0; i<n; i++){
				temp=0;
				x=num[i];
				while(x!=0){
					count[x]++;
					sort_count[x]+=temp;
					if(count[x]==k)	{
						if(min>sort_count[x]) {
							min=sort_count[x];
							break;
						}		
					}
					temp++;
					x/=2;
				}
			}
			System.out.printf("%d",min);    
		}
	}
}