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 (2.7.1) (judge verdict: Accepted)

#input of int(split by space)
def get_i()
  return gets.chomp.split(" ").map(&:to_i)
end
#input of float(split by space)
def get_f()
  return gets.chomp.split(" ").map(&:to_f)
end
#input of string(split by space)
def get()
  return gets.chomp.split(" ")
end
#input of string(split per one character)
def get_nsp()
  return gets.chomp.split("")
end
#yes or no decision
def yn_judge(bool,y="Yes",n="No")
  return bool ? y : n 
end
#create of array
def array(size1,init=nil,size2=-1)
  if size2==-1
      return Array.new(size1){init}
  else
      return Array.new(size2){Array.new(size1){init}}
  end
end

def max(a,b)
    return a>b ? a : b
end

def min(a,b)
    return a>b ? b : a
end

INF=Float::INFINITY

N=gets.to_i
a=array(N)
N.times do|i|
    a[i]=gets.to_i
end
a.sort!{|x,y| y<=>x}
if N==3
  puts max(a[0]-a[2]+a[1]-a[2],a[0]-a[1]+a[0]-a[2])
  exit
end
ans=0
if N.odd?
    ans=max(ans,(a[0..N/2-1].inject(:+)*2-(a[N/2..N-1].inject(:+)*2-a[N/2]-a[N/2+1])).abs)
else
    ans=max(ans,((a[0..N/2-1].inject(:+)*2-a[N/2-1])-(a[N/2..N-1].inject(:+)*2-a[N/2])).abs)
end
a.reverse!
if N.odd?
    ans=max(ans,(a[0..N/2-1].inject(:+)*2-(a[N/2..N-1].inject(:+)*2-a[N/2]-a[N/2+1])).abs)
else
    ans=max(ans,((a[0..N/2-1].inject(:+)*2-a[N/2-1])-(a[N/2..N-1].inject(:+)*2-a[N/2])).abs)
end
puts ans