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<vector>
#include<queue>
#include <string>
#include <sstream>
#include <complex>
#include <cmath>
#include <algorithm>
using namespace std;



int main() {
  long n, k;
  cin  >> n >> k;

  vector<long> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];/* code */
  }

  //最初座標0にいる






  long tmp = 0;
  for (int i = 1; i < k; ++i) { //開区間[0 - k)
    tmp += (a[i] - a[i - 1]); //距離
  }

  long tmp2=tmp;

  //両方共負
  if (a[0] < 0 && a[k - 1] < 0)tmp2 += -a[k - 1];
  //両方共正
  if (a[0] > 0 && a[k - 1] > 0)tmp2 += a[0];
  //両方正負
  if (a[0] < 0 && a[k - 1] > 0)tmp2 += min(abs(a[0]), abs(a[k - 1]) );


  cerr << "tmp"<< tmp << endl;
  cerr << tmp2 << endl;
  long minsum = tmp2;
  

  for (int i = 2; i + k - 2  < n; ++i) {
    tmp -= (a[i - 1] - a[i - 2]) ;
    tmp += (a[k + i - 2 ] - a[k + i - 3] );

    cerr << "tmp:"<< tmp << endl;
    tmp2 = tmp;

    //両方共負
    if (a[i - 1] < 0 && a[k + i - 2] < 0)tmp2 += -a[k + i - 2];
    //両方共正
    if (a[i - 1] > 0 && a[k + i - 2] > 0)tmp2 += a[i-1];
    //両方正負
    if (a[i - 1] < 0 && a[k + i - 2] > 0)tmp2 += min(abs(a[i-1]), abs(a[k + i - 2]) );


    cerr << "tmp2:"<<tmp2 << endl;

    if (tmp2 < minsum)minsum = tmp2;
  }


  cout << minsum;



  return 0;
}