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 <fstream>
#include <vector>
#include <queue>
#include <string>
#include <limits>
#include <algorithm>
#include <map>
#include <sstream>

using ll = long long;
using ull = unsigned long long;
using uint = unsigned int;
static ull tenq = 1000000000;
static ull mod = tenq + 7;

using namespace std;

int main()
{
  ll N, K;
  cin >> N >> K;
  vector<ll> xs(N);
  vector<ll> ls, rs;
  ll num_l = N;
  for (auto i = 0; i < N; i++) {
    cin >> xs[i];
    if (num_l == N && xs[i] >= 0) {
      num_l = i;
    }
    if (xs[i] < 0) xs[i] *= -1;
  }
  ll lidx = max(0LL, num_l - K + 1);
  ll res = tenq * 1000;
  // l
  if (num_l >= K) {
    res = xs[lidx];
  }
  // r
  if (num_l + K - 1 < N) {
    res = min(res, xs[num_l + K - 1]);
  }
  for (auto i = 0; lidx + i + K - 1 < N; i++) {
    res = min(res, xs[lidx + i] * 2 + xs[lidx + i + K-1]);
    res = min(res, xs[lidx + i] + xs[lidx + i + K-1] * 2);
  }

  cout << res << endl;

  return 0;
}