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>

#define INFTY 300000000;

int min(int a, int b){
  return a > b ? b : a;
}

int max(int a, int b){
  return a > b ? a : b;
}

int main(){
  int n, k;
  scanf("%d %d", &n, &k);
  
  int x[n + 1];
  int th = 0;
  for(int i = 0; i < n; i++){
    int in;
    scanf("%d", &in);
    if(in < 0){
      *(x + i) = in;
      th++;
    }else if(in == 0){
      *(x + i) = 0;
      *(x + i + 1) = 0;
      th++;
      k++;
    }else if(th == i){
      *(x + i) = 0;
      *(x + i + 1) = in;
    }else{
      *(x + i + 1) = in;
    }
  }
  
  int ans = INFTY;
  
  int *left = x + max(th - k, 0);
  int *right = th >= k ? x + min(left + k - 1 - x, n) : x + min(left + k - x, n);
  
  while(left <= x + th && right <= x + n){
    ans = min(ans, max(-*(left), *(right)) + 2 * min(-*(left), *(right)));
    if(*left == 0) break;
    left++;
    right++;
  }
  
  printf("%d", ans);
  
  return 0;
}