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: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(){
    int N, K;
    cin >> N >> K;

    int cp = 0, cm = 0;
    vector <int> m(0), p(0);

    for (int i = 0; i < N; i++){
        int x;
        cin >> x;

        if (x < 0){
            cm += 1;
            m.push_back(-1 * x);
        }else if(x == 0){
            K -= 1;
        }else{
            cp += 1;
            p.push_back(x);
        }
    }

    if (cm > 0){
        sort(m.begin(), m.end());
    }
    
    int sum, min = 1000000007;
    if (K == 0){
        min = 0;
    }else if(cm == 0){
        sum = p[K-1];
        if (min > sum){
            min = sum;
        }
    }else if (cp == 0){
        sum = m[K-1];
        if (min > sum){
            min = sum;
        }
    }else{
        if (cp >= K){
            sum = p[K-1];
            if (min > sum){
                min = sum;
            }
        }
        if (cm >= K){
            sum = m[K-1];
            if (min > sum){
                min = sum;
            }
        }
        if (K > 1){
            int count;
            if (0 <= K - 2 - cp){
                count = K - 2 - cp;
            }else{
                count = 0;
            }
            

            for (int i = count; i < cm && K - i - 1 > 0; i++){
                sum = m[i] * 2 + p[K - i - 2];
                if (min > sum){
                    min = sum;
                }
            }
            
            if (0 <= K - 2 - cm){
                count = K - 2 - cm;
            }else{
                count = 0;
            }
                        
            for (int i = count; i < cp && K - i - 1 > 0; i++){
                sum = p[i] * 2 + m[K - i - 2];
                if (min > sum){
                    min = sum;
                }
            }
        }   
    }
    
    cout << min << endl;

    
        

    return 0;
}