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>
#include <string.h>
#include <math.h>
#define min(a,b) (((a)<(b))?(a):(b))


int main()
{
	int n, k, z = -1;
	long int x[100100];
	long int l = 500000010;
	scanf("%d %d\n", &n, &k);
	for (int i = 0; i < n; i++)
	{
		scanf("%ld", &x[i]);
		if (z == -1 && x[i] > 0)
		{
			z = i;
		}
	}
	z--;
	if (z - k + 1 >= 0)
	{
		l = min(l, x[z - k + 1]);
	}
	int aaa = 0;
	for (int i = 0; i <= k; i++)
	{
		if (z + i - k + 1 >= 0 && z + i <= n - 1)
		{
			if (l > 2 * x[z + i] - x[z + i - k + 1])
			{
				l = 2 * x[z + i] - x[z + i - k + 1];
				aaa = i;
			}
		}
	}
	z++;
	if (z + k - 1 <= n - 1)
	{
		l = min(l, x[z + k - 1]);
	}
	for (int i = 1; i <= k; i++)
	{
		if (z - i + k - 1 <= n - 1 && z - 1 >= 0)
		{
			if (l > -2 * x[z - i] + x[z - i + k - 1])
			{
				l = -2 * x[z - i] + x[z - i + k - 1];
				aaa = -i;
			}
		}
	}
	printf("%ld", l);
	return 0;
}