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
a=[]

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

def abs(n)
  return -n if n<0
  return n
end

a = a.sort
bigger = a.select { |e| e > a[n/2]  }.reverse
smaller = a.select { |e| e <= a[n/2]  }.drop(1)

left = []
right = []

left[0] = a[0]
right[0] = a[0]
tmp = 1
indexb = 0
indexs = 0
index = 1


for i in 1..(n+1) do
  if tmp % 2 == 1
    left[index] = bigger[indexb] if bigger[indexb + 1] != nil
    right[index] = bigger[indexb+1] if bigger[indexb + 1] != nil
    right[index] = bigger[indexb] if bigger[indexb + 1] == nil && bigger[indexb] != nil
    index += 1 if bigger[index] != nil
    tmp += 1
    indexb += 2
  else
    right[index] = smaller[indexs] if smaller[indexs] != nil
    left[index] = smaller[indexs+1] if smaller[indexs+1] != nil
    index += 1
    tmp += 1
    indexs += 2
  end
end

tmp = left[0]
sum = 0
for i in left do
  sum += abs(tmp-i)
  tmp = i
end
tmp = right[0]
for i in right do
  sum += abs(tmp-i)
  tmp = i
end


puts sum