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>

#define ABS(x) ((x<0)?(x)*(-1):(x))

int get_origin(int l, int r, int *x)
{
    int o = (r+l)/2;
    if (r == o || l == o) return o;
    if (x[o] == 0) return o;
    if (x[o] > 0)
        r = o;
    else
        l = o;
    return get_origin(l,r,x);
}

int main(void) {
    int N,K;
    int i;
    int x[100000];
    int *p;
    unsigned int ans=0xFFFFFFFF;
    unsigned int tmp;
    int o;
    scanf("%d %d",&N,&K);
    for (i=0,p=x;i<N;i++,p++) {
        scanf("%d",p);
    }

    /* printf("origin = x[%d]:%d \n", o,x[o]); */

    if (x[0] >= 0)
    {
        ans = x[K-1];
    }
    else if (x[N-1] <= 0)
    {
        ans = x[N-K]*(-1);
    }
    else
    {
        o = get_origin(0,N-1,x);
        if (o-K <= 0)
            i = 0;
        else
            i = o-K+1;
        for(p=&x[i];i+K<=N;i++,p++)
        {
            if (*p > 0)
            {
                break;
            }
            else
            {
                if (*(p+K) >= 0)
                {
                    tmp = ABS(*p) + ABS(*(p+K-1) - *p);
                    if (tmp < ans)
                    {
                        ans = tmp;
                    }
                }
            }
        }
    }
    printf("%d\n",ans);

    return 0;
}