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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<string.h>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<math.h>
#include<queue>
#include<map>
#define ll long long
using namespace std;
ll a[100005];
int main()
{
	ll n, k, x,num=0,num1,ans=0,mi_n;
	scanf("%lld %lld", &n, &k);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
		if (a[i] < 0)
			num = i;
	}
	if (num == 0)
		printf("%lld", a[k]);
	else {
		if (num < k) {
			num = k;
			num1 = 1;
			ans = min(2 * a[num] + abs(a[num1]), 2 * abs(a[num1++]) + a[num++]);
		}
		else {
			num1 = n - k + 1;
			ans = abs(a[num1]);
		}

		while (num <= n) {
			if (a[num1] >= 0) {
				ans = min(ans, a[num]);
				break;
			}
			ans = min(min(2 * a[num] + abs(a[num1]), 2 * abs(a[num1++]) + a[num++]), ans);
		}
		printf("%lld", ans);
	}
	return 0;
}