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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class EqualizingDivision {
    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());
        ArrayList<Integer> arr = new ArrayList<>();
        st = new StringTokenizer(f.readLine());
        int[][] seenAndCount = new int[(int)(2 * (10e5))+1][2];
        for (int i = 0; i < n; i++) {
            arr.add(Integer.parseInt(st.nextToken()));
            seenAndCount[arr.get(i)][0]++;
        }
        Collections.sort(arr);
        int ans = Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            int cur = arr.get(i);
            int countDivides = 0;
            while (cur >= 0) {
                if (seenAndCount[cur][0] >= k) {
                    ans = Math.min(seenAndCount[cur][1], ans);
                }
                cur/=2;
                countDivides++;
                seenAndCount[cur][0]++;
                seenAndCount[cur][1] += countDivides;
                if (cur == 0) {
                    break;
                }
            }
        }
        System.out.println(ans);
    }
}