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.io.*;
import java.util.*;
     
public class _InVoker_ {
	static long mod = 1000000007;
	static long mod2 = 998244353;
    public static void sort(long 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);	
	}	
	private static void merge(long arr[], int start, int mid, int end) {		
		int i, j=mid+1,c=0;
		long temp[]= new long[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];
	}
	static long gcd(long a, long b) { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }
	static int gcd(int a, int b) { 
	      if (b == 0) 
	        return a; 
	      return gcd(b, a % b);  
	    }
	
	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();
		long a[]=new long[n];
		for(int i=0;i<n;i++) {
			a[i]=inp.nextLong();
		}
		sort(a,0,n-1);
		long gg=Long.MAX_VALUE;
		long preSum[]=new long[n+1];
		long postSum[]=new long[n+1];
		for(int i=0;i<n;i++) {
			preSum[i+1]=a[i]+preSum[i];
		}
		for(int i=n-1;i>=0;i--) {
			postSum[i]=a[i]+postSum[i+1];
		}
		for(int i=0,j=0;i<n;i=j) {
			long curElement=a[i];
			while(j<n && a[j]==curElement) {
				j++;
			}
			if(j-i>=k) {
				gg=0;
				break;
			}
			int need=k-(j-i);
			int before=i,after=n-j;
			long beforeCost= i*curElement - preSum[i];
			long afterCost= postSum[j]-(n-j)*curElement;
			if(before>=need) {
				gg=Math.min(gg, beforeCost-(before-need));
			}
			if(after>=need) {
				gg=Math.min(gg, afterCost -(after-need));
			}
			gg=Math.min(gg, beforeCost+afterCost-(before+after-need));
		}
		out.println(gg);
    	
    	inp.close();
    	out.close();
    }	
}