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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
int desc(const void *a, const void *b){
  return *(int *)b - *(int *)a;
}

int main(){
  int n;
  long long int a[110000];
  long long int b[110000];
  long long int ans=0;
  scanf("%d",&n);
  int i;
  for(i=0;i<n;i++){
    scanf("%lld",&a[i]);
  }
  i=0;
  int k=n-1;
  qsort(a,n, sizeof(long long int), desc);
  int p=n/2-1,q=n/2,c=0;
  while(k-i>0){
    if(c==0){
      b[q]=a[k];
      b[p]=a[i];
    }
    p--; q++;
    k--; i++;
    if(c%2==0){
      b[q]=a[i];
      b[p]=a[k];
    }
    else{
      b[q]=a[k];
      b[p]=a[i];
    }
    c++;
  }
  for(i=1;i<n;i++){
    ans+=llabs(b[i-1]-b[i]);
  }
  printf("%lld\n",ans);
  return 0;
}