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.*;
import java.util.*;

public class CF1213D2 extends PrintWriter {
	CF1213D2() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1213D2 o = new CF1213D2(); o.main(); o.flush();
	}

	static class V {
		int a, c;
		V(int a, int c) {
			this.a = a; this.c = c;
		}
	}
	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		V[] vv = new V[n * 20];
		int m = 0;
		for (int i = 0; i < n; i++) {
			int a = sc.nextInt();
			for (int c = 0; ; c++) {
				vv[m++] = new V(a, c);
				if (a == 0)
					break;
				a /= 2;
			}
		}
		Arrays.sort(vv, 0, m, (u, v) -> u.a != v.a ? u.a - v.a : u.c - v.c);
		int ans = n * 20;
		for (int i = 0, j; i < m; i = j) {
			j = i + 1;
			while (j < m && vv[j].a == vv[i].a)
				j++;
			if (j - i >= k) {
				int c = 0;
				for (int h = 0; h < k; h++)
					c += vv[i + h].c;
				ans = Math.min(ans, c);
			}
		}
		println(ans);
	}
}