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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<math.h>
using namespace std;
int a[100010];
int main()
{
    int n,k,i,j,x,t,f;
    cin >> n >> k;
    f=0;x=n;
    for(i=0;i<n;i++)
    {
        cin >> a[i];
        if(!f && a[i]>0)
        {
            x=i;
            f=1;
        }
    }
    int s=0,mi;
    if(x==0)
        cout << a[k-1] << endl;
    else if(x==n)
        cout << abs(a[n-k]) << endl;
    else {
        if(x<k)
        {
            if(abs(a[0])<a[k-1])
                s=2*abs(a[0])+a[k-1];
            else
                s=abs(a[0])+2*a[k-1];
        }
        else if(x==k)
            s=abs(a[0]);
        else
            s=2*abs(a[x-k]);
        mi=s;
        f=0;
        for(i=max(x,k);i<x+k;i++)
        {
            if(i>=n)
                break;
            if(a[i-k+1]>0)
            {
                if(a[x+k-1]<mi)
                    mi=a[x+k-1];
                break;
            }
            if(2*abs(a[i-k+1])+a[i]<mi)
                mi=abs(2*a[i-k+1])+a[i];
            if(abs(a[i-k+1])+2*a[i]<mi)
                mi=abs(a[i-k+1])+2*a[i];
        }
        cout << mi << endl;
    }
    return 0;
}