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:
// No sorceries shall previal. //
         
import java.util.Scanner; 
import java.io.PrintWriter;
import java.util.*;
import java.util.Arrays;
     
public class InVoker2 {
    	
    public static void sort(int arr[], int start, int end) {
    		
		if(start>=end)
			return;
			
		int mid=(start+end)/2;
		sort(arr,start,mid);
		sort(arr,mid+1,end);
		merge(arr,start,mid,end);	
	}
		
	public static void merge(int arr[], int start, int mid, int end) {
		
		int i, j=mid+1,c=0;
		int temp[]= new int[end-start+1];
		for(i=start;i<=mid && j<=end;c++) {
			if(arr[i]<=arr[j]) {
				temp[c]=arr[i];
				i++;
			}
			else {
				temp[c]=arr[j];
				j++;
			}
		}
		while(i<=mid) {
			temp[c]=arr[i];
			i++;c++;
		}
		while(j<=end) {
			temp[c]=arr[j];
			j++;c++;
		}
		c=0;
		for(i=start;i<=end;i++,c++)
			arr[i]=temp[c];
	}
    	
    public static void main(String args[]) {
    		
    	Scanner inp = new Scanner(System.in);
    	PrintWriter out= new PrintWriter(System.out);
    		
		int n=inp.nextInt();
		int k=inp.nextInt();
		
		int a[]= new int[n];
		for(int i=0;i<n;i++)
			a[i]= inp.nextInt();
		sort(a,0,n-1);
		long gg=Long.MAX_VALUE;
		int stepsToTemp[]=new int[200001];
		int count[]=new int[200001];
		
		for(int i=0;i<n;i++) {
			int temp=a[i];
			int j=1;
			while(temp>0) {
				count[temp]++;
				if(count[temp]>=k)
					gg=Math.min(gg, stepsToTemp[temp]);	
				temp/=2;
				stepsToTemp[temp]+=j++;
			}
		}
		out.println(gg);	
    	
    	out.close();
    	inp.close();
    					
    }
    		
}