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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int main(int argc, const char * argv[]) {
    int N = 0, K = 0;
    scanf("%d%d", &N, &K);
    
    int x[102000];
    
    for(int i = 0; i < N; i++){
        scanf("%d", &x[i]);
    }
    
    long long int sum_total[120000];
    for(int i = 0; i <= N - K; i++){
        long long int sum = 0;
        if(x[i] * x[i+K-1] < 0){
            if(abs(x[i]) >= abs(x[i+K-1])){
                sum = abs(x[i]) + abs(x[i+K-1]) * 2;
            }else{
                sum = abs(x[i]) * 2 + abs(x[i+K-1]);
            }
        }else{
            if(x[i] < 0){
                sum = abs(x[i]);
            }else{
                sum = abs(x[i+K-1]);
            }
        }
        
        sum_total[i] = sum;
    }
    
    long long int min = 100000000000;
    
    for(int i = 0; i <= N - K; i++){
        if(min > sum_total[i])
            min = sum_total[i];
    }
    printf("%lld\n", min);
    return 0;
}