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 Ruby (2.3.3) (judge verdict: Wrong Answer)

#!/usr/bin/env ruby

(n, k) = gets.chomp.split(/ /).map(&:to_i)
x = gets.chomp.split(/ /).map(&:to_i)
if (k == 1 && x.include?(0))
  puts '0'
  exit
end
k -= 1 if x.include?(0)

minus = x.select {|item| item < 0}
plus = x.select {|item| item > 0}

if plus.size > k
  plus = plus[0..(k - 1)]
end

minus.reverse!
if minus.size > k
  minus = minus[0..(k - 1)]
end

pplus = []
i = 0
plus.size.times do |j|
  pplus << (plus[j] - i)
  i = plus[j]
end

mminus = []
i = 0
minus.size.times do |j|
  mminus<< (minus[j] - i).abs
  i = minus[j]
end

# plus case

pcase = mminus.map{|m| m * 2}.reverse + pplus
mcase = mminus.reverse + pplus.map{|p| p * 2} 

#p pcase
#p mcase
min = 10 ** 8
(0..(pcase.size - k)).each do |x|
  kouho = pcase[x..(x + k - 1)].inject{|sum, i| sum + i}
  if kouho < min
    min = kouho
  end
  kouho2 = mcase[x..(x + k - 1)].inject{|sum, i| sum + i}
  if kouho2 < min
    min = kouho2
  end
end
puts min