Problem:
Score : 400 points

Problem Statement

You are given N integers; the i-th of them is A_i.
Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Constraints

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the maximum possible sum of the absolute differences between the adjacent elements after arranging the given integers in a row in any order you like.

Sample Input 1

5
6
8
1
2
3

Sample Output 1

21

When the integers are arranged as 3,8,1,6,2, the sum of the absolute differences between the adjacent elements is |3 - 8| + |8 - 1| + |1 - 6| + |6 - 2| = 21. This is the maximum possible sum.

Sample Input 2

6
3
1
4
1
5
9

Sample Output 2

25

Sample Input 3

3
5
5
1

Sample Output 3

8

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

#!/usr/bin/env ruby

n = gets.chomp.to_i
a = []
n.times do
  a << gets.chomp.to_i
end

a.sort!
newa = []
newb = []
if (a.size % 2 != 0)
  c = a.size / 2
  t = a.size / 2
  b = a.clone
#  if (a[c] - a[c - 1]).abs > (a[c] - a[c + 1]).abs
    newa << a.delete_at(c)
    t.times do
      newa << a.shift
      newa << a.pop if a.size > 0
    end
#  else
    newb << b.delete_at(c)
    t.times do
      newb << b.pop
      newb << b.shift if b.size > 0
    end
#  end
else
  c = a.size / 2 - 1
  t = a.size / 2
#  if (a[c] - a[-1]).abs > (a[c + 1] - a[0]).abs
    newa << a.delete_at(c)
#
    t.times do
      newa << a.pop
      newa << a.shift if a.size > 0
    end
#  else
    newb << b.delete_at(c + 1)
#
    t.times do
      newb << b.shift
      newb << b.pop if b.size > 0
    end
#  end


end
#p newa
res = 0
s = newa.shift
newa.each do |e|
  res += (s - e).abs
  s = e
end

res2 = 0
s = newb.shift
newb.each do |e|
  res2 += (s - e).abs
  s = e
end

puts [res, res2].max