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: Wrong Answer)

def f1(n, a)
  b = [a[0]]
  a.shift
  
  m = a.size
  (0..(m / 2 - 1)).each do |i|
    b.unshift(a[(m - 1) - (2 * i)])
    b.push(a[(m - 1) - (2 * i + 1)])
  end
  
  if n % 2 == 0 then
    b.unshift(a[0])
  end
  
  return b
end

def f2(n, a)
  b = [a[n - 1]]
  a.pop

  m = a.size
  (0..(m / 2)).each do |i|
   if i % 2 == 0 then
     b.unshift( a.shift )
     b.push( a.shift )
   else
     b.unshift( a.pop )
     b.push( a.pop )
   end
  end

  b.pop
  if n % 2 == 1 then
    b.shift
  end

  return b
end

def sum(a)
  s = 0
  
  (a.size - 1).times do |i|
    s += (a[i + 1] - a[i]).abs
  end
  
  return s  
end

n = gets.chomp.to_i
a = []
n.times{ |i| a[i] = gets.chomp.to_i }
a.sort!

b = f1(n, a.dup)
c = f2(n, a.dup)

puts [sum(b), sum(c)].max.to_s