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: Runtime Error)

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>
#define MAX(a,b) ((a)>(b)?(a):(b))
int cmp(const void *a, const void *b)
{
  return *(long*)a - *(long*)b;
}

int main()
{
  long n, i, n_half;
  long a[10000];
  long ans=0;
  long ans1, ans2;

  scanf("%ld",&n);
  for(i=0;i<n;i++) {
    scanf("%ld",&a[i]);
  }
  qsort(a, n, sizeof(long), cmp);
  if(n%2==0) {
    n_half = n/2;
    for(i=0;i<n_half-1;i++) {
      ans = ans + 2*a[n-1-i] - 2*a[i];
    }
    ans = ans + a[n_half] - a[n_half-1];
  } else {
    n_half = n/2;
    for(i=0;i<n_half-1;i++) {
      ans = ans + 2*a[n-1-i] - 2*a[i];
    }
    ans1 = ans - a[n_half-1] + a[n_half] + a[n_half+1];
    ans2 = ans - a[n_half-1] - a[n_half] + a[n_half+1];
    ans = MAX(ans1, ans2);
  }
  printf("%ld",ans);
  return 0;
}