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)

require 'pp'

n = gets.chomp!.to_i
a_n = Array.new(n)
n.times do |i|
    a_n[i] = gets.to_i
end
a_n.sort!

if n % 2 == 1
    before = a_n[n / 2] 
    a_n[n / 2] = 0
    if (before - a_n[0]).abs >= (before - a_n[-1]).abs
        i = 1
    else
        i = 0
    end
else
    t1 = a_n[n / 2 - 1] - a_n[-1]
    t2 = a_n[n / 2] - a_n[0]
    if t1.abs >= t2.abs
        before = a_n[n / 2 - 1]
        a_n[n / 2 - 1] = 0
        i = 0
    else
        before = a_n[n / 2]
        a_n[n / 2] = 0
        i = 1
    end
end

ans = 0
while a_n.empty?.!
    x = if i % 2 == 0
            a_n.pop
        else
            a_n.shift
        end

    if x == 0
        break
    end

    #pp x
    #pp before
    #pp a_n

    ans += (x - before).abs
    before = x
    i += 1
end

puts ans