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.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        long[] num = new long[200010];
        long[] sum = new long[200010];
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();//
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
        }
        Arrays.sort(arr);
        //找到一共有多少个数可以变成arr[i]以及他们所需要的次数
        long ans=Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            int x = arr[i];
            int count = 0;
            while (x > 0) {
                num[x]++;
                sum[x] += count++;
                if (num[x] == k) {
                    ans = Math.min(ans, sum[x]);
                }
                x /= 2;
            }
        }
        System.out.println(ans);
    }
}