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 <stdlib.h>
#include <string.h>
#include <math.h>
#include <stdio.h>
int compare(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}
int main()
{    int n,i,nmin=0,j=0,k=0;
long long ans=0,a[100004],b[100004],tmp1,tmp2,large[50001],small[50001];
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%lld",&a[i]);
    }
    qsort(a,n,sizeof(long long),compare);
    for(i=0;i<n;i++){
        if(a[i]==a[i+1]){
        nmin++;
        }
        small[i]=a[i];
        large[i]=a[n-i-1];
    }
    tmp1=a[nmin+1];
    for(i=nmin;i>=0;i--){
        if(i==0){a[0]=tmp1;
    }a[i+1]=a[i];
}
    for(i=0;i<n-1;i++){
        printf("%lld\n",b[i]);
        ans+=abs(b[i]-b[i+1]);
    }
    
    printf("%lld",ans);

    return 0;
}