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.PriorityQueue;
import java.util.Scanner;

public class d {	
	public static void main(String[] args) {
		Scanner stdin = new Scanner(System.in);
		
		int n = stdin.nextInt();
		int k = stdin.nextInt();
		
		PriorityQueue<Integer>[] arr = new PriorityQueue[200_001];
		for (int i = 0; i < 200_001; i ++) arr[i] = new PriorityQueue<>();
		
		for (int i = 0; i < n; i ++) {
			int in = stdin.nextInt();
			arr[in].add(0);
			int count = 0;
			while (in >= 1) {
				in /= 2;
				arr[in].add(++count);
			}
		}
		
//		for (int i = 1; i < 3; i ++) for (int j = 0; j < arr[i].size(); j ++) System.out.println(arr[i].poll());
		
		int min = 1_000_000_000;
		
		for (int i = 0; i < 200_001; i++) {
			int taken = 0, sum = 0;
			while (!arr[i].isEmpty() && taken < k) {
				sum += arr[i].poll();
				taken ++;
				if (taken == k) min = Math.min(min, sum);
			}
		}
		
		System.out.println(min);
	}
}