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 <string>
#include <math.h>
#include <vector>
#include<stdio.h>
#include <functional>
#include<algorithm>
#define PI 3.14159265
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n)  FOR(i,0,n)
#define all(x) (x).begin(),(x).end()
typedef long long ll;
#define SZ(x) ((int)(x).size())

using namespace std;



int main() {
	int n, k,ans;
	cin >> n >> k;
	int x[100000];
	int s = -1; 
	ans = 0;
	int m;
	REP(i, n) {
		cin >> x[i];
		if (s == -1 && x[i] >= 0) {
			s = i;
		}
	}
	if (x[n - 1] >= 0) {
		if (s - k < 0) {
			REP(i, s) {
				if (i == 0) {
					m = min(abs(x[i + k - 1]), abs(x[i]));
					ans = m + x[i + k - 1] - x[i];
				}
				if (i != 0 && i + k - 1 <= n - 1) {
					m = min(abs(x[i + k - 1]), abs(x[i]));
					ans = min(m + x[i + k - 1] - x[i], ans);
				}
			}
		}
		else {
			FOR(i, s - k + 1, s) {
				if (i == s - k + 1) {
					m = min(abs(x[i + k - 1]), abs(x[i]));
					ans = m + x[i + k - 1] - x[i];
				}
				if (i != s - k + 1 && i + k - 1 <= n - 1) {
					m = min(abs(x[i + k - 1]), abs(x[i]));
					ans = min(m + x[i + k - 1] - x[i], ans);
				}
			}
		}

		if (s >= k) {
			ans = min(ans, x[s - k]);
		}
		if (s + k - 1 <= n - 1) {
			ans = min(ans, x[s + k - 1]);
		}
	}
	else {
		ans = x[n - k];
	}
	cout << ans << endl;

	return 0;
}