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: Accepted)

#include<stdio.h>
#include<stdlib.h>
int main(){
	int n,k;
	scanf("%d %d",&n,&k);
	long x[n],ans;
	int i,j;
	for(i=0;i<n;i++)scanf("%ld",&x[i]);
	for(i=0;i<=n-k;i++){
		if(i==0)ans=abs(x[i])+abs(x[i+k-1]-x[i]);
		if(ans>abs(x[i])+abs(x[i+k-1]-x[i]))ans=abs(x[i])+abs(x[i+k-1]-x[i]);
		if(ans>abs(x[i+k-1])+abs(x[i+k-1]-x[i]))ans=abs(x[i+k-1])+abs(x[i+k-1]-x[i]);
	}
	printf("%ld\n",ans);
	//for(i=0;i<n;i++)printf("%ld,",x[i]);
	return 0;
}