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>

int n,k,x[200000],y,border = 20000;
long long an = 9000000000000000000;

int main() {

    scanf("%d %d",&n,&k);
    for(y=1;y<=n;y++){
        scanf("%d", &x[y]);
        if(x[y]>0 && border == 20000 ) border = y-1;
        if(x[y] == 0){
            y--;
            n--;
            k--;
        }
    }
    if(x[1]>=0){
        printf("%d\n",x[k]);
        return 0;
    }
    if(border >= k) an = x[border-k+1]*-1;
    if(n-border >= k && x[border+k] < an) an = x[border+k];

    k--;

    if(x[n]<=0){
        printf("%d\n",x[n-k]*-1);
        return 0;
    }


    for(y=1;x[y]<0;y++){
        if( y+k == n || x[y+k] < 0 && y+k != border || y+k > n) break;
        if( x[y]*-2 + x[y+k] < an ) an = x[y]*-2 + x[y+k];
        if( x[y]*-1 + x[y+k]*2 < an ) an = x[y]*-1 + x[y+k]*2;
    }
    printf("%lld\n",an);

    return 0;
}