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>

long minf(long a,long b){
  if(a<b){
    return a;
  }else{
    return b;
  }
}
int main(){
  long n;
  long k;
  long a[100010] = {};
  long plus = 0;
  long minus = 0;
  long zero = 0;
  long min = 10000000007;
  long i = 0;
  scanf("%ld %ld",&n,&k);
  for(i=0;i<n;i++){
    scanf("%ld",&a[i]);
  }
  
  for(i=0;i+k-1<n;i++){
    if(a[i+k-1] >= 0 && a[i] >= 0){
      if(labs(a[i+k-1]) < min){
        min = labs(a[i+k-1]);
        //printf("#");
      }
    }else if(a[i+k-1] <= 0 && a[i] <= 0){
      if(labs(a[i]) < min){
        min = labs(a[i]);
      }
    }else{
      if(-a[i] <= a[i+k-1]){
        if(min > -2*a[i] + a[i+k-1]){
          min = -2*a[i] + a[i+k-1];
        }
      }else{
        if(min > -a[i] + 2*a[i+k-1]){
          min = -a[i] + 2*a[i+k-1];
        }
      }
    }
  }
  printf("%ld",min);
  return 0;
}