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 main(void){
    int flg = 0;
    int n,i,j,tmp,ans;
    scanf("%d",&n);
    int a[n];
    for(i=0;i<n;i++){
        a[i] = 0;
        scanf("%d",&a[i]);
    }
    for (i=0; i<n; ++i) {
        for (j=i+1; j<n; ++j) {
            if(a[i] == a[j]){
                flg = 1;
            }
            if (a[i] < a[j]) {
                tmp =  a[i];
                a[i] = a[j];
                a[j] = tmp;
            }
        }
    }

    for(i=0;i<n;i++){
        if(abs(a[i]) >=abs(a[(n-1)-i])){
            ans += abs(a[i]) - abs(a[(n-1)-i]);
        }else{
            ans +=  abs(a[(n-1)-i]) - abs(a[i]);
        }
    }
    if(flg){
        printf("%d",ans);
    }else{
        printf("%d",ans-1);
    }
}