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: Runtime Error)

#include <stdio.h>

int main(void){
	int n,k;
	scanf("%d %d",&n,&k);
	int x[n];
	int m[10^5]={};
	int p[10^5]={};
	int a=1,b=1;
	int i;
	int minus = 0;
	int plus = 0;
	for(i=0;i<n;i++){
		scanf("%d",&x[i]);
			if(x[i]==0){		//もしも座標0にろうそくがあったならば
				k--;
			}
		if(x[i]<0){ //minusの座標に置かれているろうそくの数をカウントする
			minus++;
		}
		if(x[i]>0){
			p[b] = x[i];//0からb本プラス側に火をつけるのにかかる時間はp[b]
			b++;
			plus++;
		}
	}
	//printf("plusの本数は%d\n",plus);
	//printf("minusの本数は%d\n",minus);
	//原点からプラス座標にn本火をつけるのにかかる時間をp[n]とする。
	//原点からマイナス座標にn本火をつけるのにかかる時間をm[n]とする。
	for(i=minus;i>0;i--){
		m[a]=-(x[i-1]);
		a++;
	}
	//printf("0からプラス座標に2本火をつけるのにかかる時間は%d\n",p[2]);
	//printf("0からマイナス座標に1本火をつけるのにかかる時間は%d\n",m[1]);
	//あとは原点から左にi本右からk-i本火をつけると考える
	long ans = 2*100000000;
	long temp;
	long temp2;
	if(k==0){
		ans=0;
		printf("%ld",ans);
		return 0;
	}
	for(i=0;i<k;i++){
		if( (k-i)>plus){
			continue;
		}
		if( i>minus){
			continue;
		}
		temp = 2*m[i]+p[k-i];
		temp2= m[i]+2*p[k-i];
		if( ans > temp){
			ans = temp;
		}
		if(ans > temp2){
			ans =temp2;
		}
	}
	printf("%ld",ans);
	return 0;
}