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>
#include <math.h>
int llassort(const void *a, const void *b){return *(long long*)b-*(long long*)a;}
int main(void){
    int n;
    long long l,r,ans=0;
    scanf("%d",&n);
    long long a[n];
    for (int i=0; i<n; i++){
        scanf("%lld",&a[i]);
    }
    qsort (a,n,sizeof(long long),llassort); // long long 昇順
    for (int i=0; i<n; i++){
        if (i==0){
            l=a[i];
            r=a[i];
        } else if (i%2==1){
            if (llabs(a[n-i/2-1]-l)>=llabs(a[n-i/2-1]-r)){
                ans+=llabs(a[n-i/2-1]-l);
                l=a[n-i/2-1];
            } else {
                ans+=llabs(a[n-i/2-1]-r);
                r=a[n-i/2-1];
            }
        } else {
            if (llabs(a[i/2]-l)>=llabs(a[i/2]-r)){
                ans+=llabs(a[i/2]-l);
                l=a[i/2];
            } else {
                ans+=llabs(a[i/2]-r);
                r=a[i/2];
            }
        }
    }
    printf("%lld\n",ans);
	return 0 ;
}