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
import java.io.*;
import java.util.*;
public class F
{
    private static final long INF=Long.MAX_VALUE;
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
 
        int i,N;
 
        String s[]=br.readLine().trim().split(" ");
        N=Integer.parseInt(s[0]);
        int K=Integer.parseInt(s[1]);
 
        long a[]=new long[N];
        s=br.readLine().trim().split(" ");
        for(i=0;i<N;i++)
            a[i]=Integer.parseInt(s[i]);
        HashSet<Long> set=new HashSet<>();
        HashMap<Long,Integer> map=new HashMap<>();
        
        for(i=0;i<N;i++)
        {
            set.add(a[i]);
            map.put(a[i],map.getOrDefault(a[i],0)+1);
        }
        
        N=set.size(); i=0;
        Arrays.fill(a,0);
        for(long tmp:set) a[i++]=tmp;
        Arrays.sort(a,0,N);
        
        long prefix[]=new long[N+1];
        int count[]=new int[N+1];
        for(i=1;i<=N;i++) prefix[i]=prefix[i-1]+a[i-1]*map.get(a[i-1]);
        for(i=1;i<=N;i++) count[i]=count[i-1]+map.get(a[i-1]);
        
        long ans=INF;
        
        for(i=0;i<N;i++)
        {
            int needed=K-map.get(a[i]);
            long lCount=0,rCount=0;
            int lExist=count[i],rExist=count[N]-count[i+1];
 
            if(needed<=0) {ans=0;break;}
            
            if(lExist>0) lCount=lExist*(a[i]-1)-prefix[i];
            if(rExist>0) rCount=prefix[N]-prefix[i+1]-(rExist)*(a[i]+1);
 
            if(lExist>=needed) ans=Math.min(ans,lCount+needed);
            if(rExist>=needed) ans=Math.min(ans,rCount+needed);
            ans=Math.min(ans,lCount+rCount+needed);
        }
        System.out.println(ans);
    }
}