Problem:
Score : 300 points

Problem Statement

There are N candles placed on a number line.
The i-th candle from the left is placed on coordinate x_i.
Here, x_1 < x_2 < ... < x_N holds.

Initially, no candles are burning.
Snuke decides to light K of the N candles.

Now, he is at coordinate 0.
He can move left and right along the line with speed 1.
He can also light a candle when he is at the same position as the candle, in negligible time.

Find the minimum time required to light K candles.

Constraints

1 \leq N \leq 10^5

1 \leq K \leq N

x_i is an integer.

|x_i| \leq 10^8

x_1 < x_2 < ... < x_N

Input

Input is given from Standard Input in the following format:

N K
x_1 x_2 ... x_N

Output

Print the minimum time required to light K candles.

Sample Input 1

5 3
-30 -10 10 20 50

Sample Output 1

40

He should move and light candles as follows:

Move from coordinate 0 to -10.

Light the second candle from the left.

Move from coordinate -10 to 10.

Light the third candle from the left.

Move from coordinate 10 to 20.

Light the fourth candle from the left.

Sample Input 2

3 2
10 20 30

Sample Output 2

20

Sample Input 3

1 1
0

Sample Output 3

0

There may be a candle placed at coordinate 0.

Sample Input 4

8 5
-9 -7 -4 -3 1 2 3 4

Sample Output 4

10

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        final int N = sc.nextInt();
        int K = sc.nextInt();
        int[] l = new int[N+1];
        int a = N+1;
        boolean f = false;
        for(int i = 1;i <= N;i++){
            l[i] = sc.nextInt();
            if(l[i] >= 0 && !f){
                a = i;
                f = true;
            }
        }
        int mi = 0;
        int Mi = 0;
        if(a-1 >= K){
            Mi = a-1;
            mi = a-K;
        }else{
            mi = 1;
            Mi = K;
        }
        long ans = Long.MAX_VALUE;
        while(mi <= a && Mi <= N){
            if(Mi < a){
                ans = Math.min(ans, -l[mi]);
            }else if(mi == a){
                ans = Math.min(ans, l[Mi]);
            }else{
                long c = Math.min(l[Mi], -l[mi]);
                long cc = Math.max(l[Mi], -l[mi]);
                ans = Math.min(ans, c*2+cc);
            }
            mi++;
            Mi++;
        }
        System.out.println(ans);
    }
}