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 Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int k=sc.nextInt();
		int a[]=new int[n];
		int num[]=new int[200001];
		int cnt[]=new int[200001];
		for(int i=0;i<n;i++) {
			a[i]=sc.nextInt();
		}
		Arrays.sort(a);
		int ans=Integer.MAX_VALUE;
		for(int i=0; i<n; i++)
	    {
	        int temp=0;
	        int x=a[i];
	        while(x!=0)
	        {
	            cnt[x]++;
	            num[x]+=temp;
	            if(cnt[x]==k) {
	                ans=Math.min(ans,num[x]);
	            }
	            temp++;
	            x/=2;
	        }
	    }
		System.out.println(ans);
	}
}