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 <string>
#include <algorithm>
#include <vector>
#include <utility>
#include <cstring>
#include <cmath>
#define INF (long)1e9
const long MOD = 1e9 + 7;
const long dx[] = {1, -1, 0, 0};
const long dy[] = {0, 0, 1, -1};
using namespace std;
typedef long long ll;
int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	long n, k;
	cin >> n >> k;
	long A[n + 1];
	for(long i = 0; i < n; i++)
		cin >> A[i];
	A[n] = INF;
	vector<long> d;
	for(long i = 0; i < n; i++)
	{
		if(A[i + k - 1] == INF)
			break;
		if(A[i] == A[i + k - 1])
			d.push_back(A[i]);
		else if(A[i] < 0 && A[i + k - 1] < 0 || A[i] > 0 && A[i + k - 1])
			d.push_back(max(abs(A[i]), abs(A[i + k - 1])));
		else if(A[i] < 0 && A[i + k - 1] > 0)
		{
			d.push_back(abs(A[i]) + abs(A[i + k - 1]));
			d[i] += min(abs(A[i]), abs(A[i + k - 1]));
		}
	}
	long ans = INF;
	for(long i = 0; i < d.size(); i++)
	{
		if(d[i] < ans)
			ans = d[i];
	}
	cout << ans << endl;
	return 0;
}