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 <algorithm>
#include <cassert>
#include <climits>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
using namespace std;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, K;
  cin >> N >> K;
  vector<int> xs;
  for (int i = 0; i < N; ++i) {
    int xi;
    cin >> xi;
    xs.push_back(xi);
  }

  int ans = INT_MAX;
  for (int i = 0; i < N - K + 1; ++i) {
    int min_val = xs[i];
    int max_val = xs[i + K - 1];
    if (min_val < 0 && max_val < 0) {
      ans = min(ans, abs(min_val));
    } else if (min_val > 0 && max_val > 0) {
      ans = min(ans, max_val);
    } else {
      if (abs(min_val) < max_val) {
        ans = min(ans, 2 * abs(min_val) + max_val);
      } else {
        ans = min(ans, abs(min_val) + 2 * max_val);
      }
    }
  }

  cout << ans << endl;
  return 0;
}