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)

n = gets.chomp.to_i
list = n.times.map { gets.chomp.to_i }.sort

if n.odd?
  p_idx = (n / 2)
  piv = list[p_idx]
  if (piv - list[0]).abs > (piv - list[-1]).abs
    front = 1
  else
    front = 0
  end
else
  piv1 = list[(n / 2) - 1]
  piv2 = list[(n / 2)]
  if (piv1 - list[-1]).abs > (piv2 - list[0]).abs
    piv = piv1
    p_idx = (n / 2) - 1
    front = 0
  else
    piv = piv2
    p_idx = n / 2
    front = 1
  end
end

bef = piv
sum = 0
f_idx = 0
b_idx = n - 1
idx = front == 1 ? f_idx : b_idx
while p_idx != idx
  sum += (bef - list[idx]).abs
  if front == 1
    f_idx += 1
  else
    b_idx -= 1
  end
  front = 1 - front
  bef = list[idx]
  idx = front == 1 ? f_idx : b_idx
end
puts sum