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>
#define inf 1<<30
#define min(a,b) a<b?a:b
int needTime(int x[], int start, int end);
int main(void)
{
    int n,k;
    scanf("%d %d",&n,&k);
    int x[n],y[n];
    for(int i = 0; i < n; i++)
    {
        scanf("%d",&x[i]);
    }
    for(int i = 0; i < n; i++)
    {
        y[i] = inf;
    }

    int tmp = 0;
    while(tmp <= n - k)
    {
        y[tmp] = needTime(x,tmp,k);
        tmp++;
    }

    int ans = inf;
    for(int i = 0; i < n; i++)
    {
        ans = min(y[i],ans);
    }
    printf("%d\n",ans);
}

int needTime(int x[], int start, int k)
{
    if(x[start] >= 0)
    {
        return x[start + k - 1];
    }
    if(x[start + k - 1] <= 0)
    {
        return x[start];
    }

    return min(abs(x[start]) * 2 + x[start + k - 1], abs(x[start]) + x[start + k - 1] * 2);
}