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<algorithm>
using namespace std;
//WA
int main() {
	int n, k; cin >> n >> k;
	int x[114514], flag = -1;
	for (int i = 0; i < n; i++) {
		cin >> x[i];
		if (x[i] == 0) {
			flag = i;
			k--;
		}
		else if (flag == -1 && x[i] > 0) {
			n++; x[i + 1] = x[i];
			x[i] = 0; flag = i++;
		}
	}
	if (k == 0)cout << "0\n";
	else if (flag == -1)cout << -x[k - 1] << endl;
	else if (flag == 0)cout << x[k] << endl;
	else {
		int ans, p = flag + 1, m = flag - 1;
		for (int i = 0; i < k; i++) {
			if (p == n)m--;
			else if (m == -1)p++;
			else if (x[p] - x[p - 1] < (x[m] - x[m - 1]) * 2)p++;
			else if (x[p] - x[p - 1] > (x[m] - x[m - 1]) * 2)m--;
			else {
				//
			}
		}
		p--; m++;
		ans = x[p] - x[m] * 2;
		p = flag + 1; m = flag - 1;
		for (int i = 0; i < k; i++) {
			if (p == n)m--;
			else if (m == -1)p++;
			else if ((x[p] - x[p - 1]) * 2 < x[m] - x[m - 1])p++;
			else if ((x[p] - x[p - 1]) * 2 > x[m] - x[m - 1])m--;
			else {
				//
			}
		}
		p--; m++;
		ans = min(x[p] * 2 - x[m], ans);
		cout << ans << endl;
	}
}