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>
#define ll long long int
ll mima,plma;
ll answer=999999999999;
ll d(int n,int k,ll pl[],ll mi[],ll dp,int now,ll ans,ll p,ll m){
  ll tasu;
  if(dp==k){
    if(ans<answer){
      answer=ans;
    }
  }
  else{
    if(mima>m){
      if(now>0)tasu=pl[p]-mi[m+1];
      else if(now<0)tasu=mi[m]-mi[m+1];
      else tasu=mi[m+1];
      d(n,k,pl,mi,dp++,mi[m+1],ans+=tasu,p,m++);
    }
    if(plma>p){
      if(now>0)tasu=pl[p]-pl[p+1];
      else if(now<0)tasu=pl[p+1]-mi[m];
      else tasu=pl[p+1];
      d(n,k,pl,mi,dp++,pl[p+1],ans+=tasu,p++,m);
    }
  }
  return 0;
}
int main(){
  int n,k;
  int i,j,l;
  ll x[110000],mi[110000],pl[110000];
  mi[0]=0;
  pl[0]=0;
  scanf("%d%d",&n,&k);
  for(i=0;i<n;i++){
    scanf("%d",&x[i]);
  }
  for(i=0;i<n;i++){
    if(x[i]>=0){
      l=1;
      for(j=i;j>=0;j--){
        mi[l]=x[j];
        l++;
      }
      mima=l;
      l=1;
      for(j=i;j<n;j++){
        pl[l]=x[j];
        l++;
      }
      plma=l;
    }
  }
  d(n,k,pl,mi,0,0,0,0,0);
  printf("%lld",answer);
  return 0;
}