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.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

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 < a.length; i++) {
			a[i] = sc.nextInt();
		}
		System.out.println(solve(a, k));

		sc.close();
	}

	static int solve(int[] a, int k) {
		Map<Integer, List<Integer>> valueToOperationNums = new HashMap<>();

		for (int ai : a) {
			int operationNum = 0;
			while (true) {
				if (!valueToOperationNums.containsKey(ai)) {
					valueToOperationNums.put(ai, new ArrayList<>());
				}
				valueToOperationNums.get(ai).add(operationNum);

				if (ai == 0) {
					break;
				}

				operationNum++;
				ai /= 2;
			}
		}

		return valueToOperationNums.values().stream().filter(operationNums -> operationNums.size() >= k)
				.mapToInt(operationNums -> operationNums.stream().sorted().limit(k).mapToInt(x -> x).sum()).min()
				.getAsInt();
	}
}