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];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        Arrays.sort(a);

        int min = Integer.MAX_VALUE;
        for (int i = 0; i <= a[n-1]; i++) {
            int num = solve(a, i, k);
            min = Math.min(min, num);
        }

        System.out.println(min);
    }

    private static int solve(int[] a, int target, int k) {
        int ret = 0;
        int count = 0;
        for (int i = 0; i < a.length; i++) {
            int w = a[i];
            if (w < target) {
                continue;
            }
            int opCount = 0;
            while (w > target) {
                opCount++;
                w /= 2;
            }
            if (target == w) {
                ret += opCount;
                count++;
            }
            if (count >= k) {
                break;
            }
        }

        return count >= k ? ret : Integer.MAX_VALUE;
    }
}