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 <stdlib.h>
void quick(int,int,int *);

int main(){
    int n,i,j,half,max=0,sum=0,temp;

    scanf("%d",&n);

    int p[n],q[n];
    for(i=0;i<n;i++){
        scanf("%d",&p[i]);
        //printf("p[%d]=%d\n",i,p[i]);
    }

    quick(0,n-1,p);
    for(i=0;i<n;i++){
        //printf("p[%d]=%d\n\n",i,p[i]);
    }

    for(i=0;i<=n/2;i++){
        q[i*2]=p[i];
        if(i*2+1<n){
            q[i*2+1]=p[n-1-i];
        }
    }
    
    for(j=0;j<n;j++){
        for(i=0;i<n-1;i++){
            //printf("q[%d]=%d\n",i,q[i]);
            sum+=abs(q[i]-q[i+1]);
        }
        //printf("q[%d]=%d\n\n",n-1,q[n-1]);
        if(sum>max){
            max=sum;
        }
       temp=q[0];
        for(i=0;i<n-1;i++){
            q[i]=q[i+1];
        }
        q[n-1]=temp;
        sum=0;
    }
    printf("%d",max);

    return 0;
}

void quick(int left,int right, int *p){
    int s,i,j,temp;

    if(left<right){
        s=p[left];
        i=left+1;
        j=right;
        
        while(1){
            while(p[i]<s) i++;
            while(p[j]>s) j--;
            if(i>=j) break;
            temp=p[i];
            p[i]=p[j];
            p[j]=temp;
        }
        p[left]=p[j];
        p[j]=s;

        quick(left,j,p);
        quick(i,right,p);
    }
}