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>
#include <string.h>
#include <time.h>
#include <math.h>
#include <limits.h>
#include <assert.h>
#define FOR(i,a,n) for(i=(a);i<(n);i++)
#define swap(type,a,b) do{type t=a;a=b;b=t;}while(0)
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))
#define ll long long
#define INF 100000000
#define MOD 1000000007
#define SIZE 100005
#define LOCAL
#ifdef LOCAL
    #define DEBUG(...) printf(" "__VA_ARGS__)
#else
    #define DEBUG(...)
#endif /* LOCAL */
int comp(const void* a,const void* b){
	return *(int*)a-*(int*)b;
}
int x[100000];
int main(void)
{
	int n,k,i;
	ll n1,n2,t,res=1000000000000000LL;
	scanf("%d%d",&n,&k);
	FOR(i,0,n) scanf("%d",&x[i]);
	FOR(i,0,n-k+1){
		if(x[i]>=0&&x[i+k-1]>=0)  t=MAX(x[i],x[i+k-1]);
		else if(x[i]<=0&&x[i+k-1]<=0)  t=-MIN(x[i],x[i+k-1]);
		else{
			n1=abs(x[i]),n2=abs(x[i+k-1]);
			if(n1>n2) t=n1+2*n2;
			else t=2*n1+n2;
		}
		res=MIN(res,t);
	}
	printf("%lld\n",res);
	return 0;
}