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 main(void){
		int n,k;
		scanf("%d %d",&n,&k);
		long x[100000];
		long m[100000]={};
		long p[100000]={};
		m[0]=0;
		p[0]=0;
		int a=1,b=1;
		int i;
		int minus = 0;//マイナス側に置かれているろうそくの数
		int plus = 0; //プラス側に置かれているろうそくの数
		for(i=0;i<n;i++){
			scanf("%ld",&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本のろうそくを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*1000000000;
		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;
	}