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>
int a[100005];
void quicksort(int list[], int left, int right)
{
    int i, last;
    int temp;

    if (left >= right)
        return;

    last = left;
    for (i = left + 1; i <= right; i++) {
        if (list[i] < list[left]) {
            last++;
            temp = list[last];
            list[last] = list[i];
            list[i] = temp;
        }
    }

    temp = list[left];
    list[left] = list[last];
    list[last] = temp;

    quicksort(list, left, last - 1);
    quicksort(list, last + 1, right);
}
int main()
{
    int n;
    scanf("%d", &n);
    int i;
    for (i = 0; i < n; i++)
        scanf("%d", &a[i]);
    int x;
    quicksort(a, 0, n - 1);
    double ans = 0;
    int max = a[0], min = a[0];
    int left = 1, right = n - 1;
    i = 0;
    while (left <= right)
    {
        switch (i % 4)
        {
        case 0:
        {
            ans += (double)(a[right] - min);
            min = a[right];
            right--;
            i++;
            break;
        }
        case 1:
        {
            ans += (double)(a[right] - max);
            max = a[right];
            right--;
            i++;
            break;
        }
        case 2:
        {
            ans += (double)(min - a[left]);
            min = a[left];
            left++;
            i++;
            break;
        }
        case 3:
        {
            ans += (double)(max - a[left]);
            max = a[left];
            left++;
            i++;
            break;
        }
        }
    }
    printf("%.0lf\n", ans);
    return 0;
}