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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>

int main(){
  int n,a[100000];
  int temp,ans=0;
  int i,j;
  scanf("%d", &n);
  for(i=0;i<n;i++){
    scanf("%d", &a[i]);
  }

  for(i=0;i<n;i++){
    for(j=n-1;j>i;j--){
      if(a[j]<a[j-1]){
	temp=a[j];
	a[j]=a[j-1];
	a[j-1]=temp;
      }
    }
  }
  if(n%2==0){
    for(i=0;i<n/2-1;i++){
      ans=ans-a[i]*2+a[n-i-1]*2;
    }
    ans=ans+a[n/2]-a[n/2-1];
  }  
  else{
    for(i=0;i<n/2;i++){
      ans=ans-a[i]*2+a[n-i-1]*2;
    }
    if(a[n/2+1]-a[n/2]>a[n/2]-a[n/2-1]) ans=ans-a[n/2]+a[n/2-1];
    else ans=ans-a[n/2+1]+a[n/2];
  }  
  printf("%d\n",ans);
  return 0;
}