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.*;
public class equalizingarraybydivision {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int k=sc.nextInt();
ArrayList<Integer> ar[]=new ArrayList[200001];
for(int i=0;i<=200000;i++)
{
	ar[i]=new ArrayList<>();

}
for(int i=0;i<n;i++)
{
int x=sc.nextInt();
int ct=0;
while(x!=0)
{
	ar[x].add(ct++);
	x=x/2;
	
}

}
int min=(int)1e9;
for(int i=0;i<=200000;i++)
{
	Collections.sort(ar[i]);
	if(ar[i].size()<k)
		continue;
	int s=0;
	for(int j=0;j<k;j++)
	{
		s+=ar[i].get(j);
	}
		min=Math.min(min,s);
}
System.out.println(min);
	}

}