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 <vector>
using namespace std;
int main(){
	int n,k;cin >> n >> k;
	vector<int> cv;
	for(int i = 0; i < n;i++){
		int c; cin >> c;
		cv.push_back(c);
	}

	int cost = abs(cv[k-1]- cv[0]);
	if(cv[0] < 0 && cv[k-1] > 0)cost += min(abs(cv[0]),abs(cv[k-1]));
	int ans = cost;

	for(int i = k; i < n; i++){
			int n_cost = abs(cv[i-k+1]- cv[i]);
			if(cv[i-k+1] < 0 && cv[i] > 0)n_cost += min(abs(cv[i-k+1]),abs(cv[i]));
			ans = min(n_cost, ans);
	}
	if(cv[0] >= 0)ans = cv[k-1];
	if(cv[n-1] < 0)ans = abs(cv[n-k]);

	cout << ans << endl;
}