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>
void sortm(int number[1000000],int n){
 int i, j, tmp;
 
  for (i=0; i<n;i++) {
    for (j=i+1; j<n;j++) {
      if (number[i] < number[j]) {
        tmp =  number[i];
        number[i] = number[j];
        number[j] = tmp;
      }
    }
  }
}

int ab(int a,int b){
    if(a>b)
    return(a-b);
    else
    return(b-a);
}


int main(){
    int A[1000000],i,N,B[1000000],sum=0;
    
    scanf("%d",&N);
    
    for(i=0;i<N;i++)
    scanf("%d",&A[i]);
    
    sortm(A,N);
   
   for(i=0;i<N/2;i++){
   B[2*i+1]=A[i];
   B[0]=A[N/2];
   }
   
    if(N%2==1){
   B[N-1]=A[1+N/2];
    for(i=1;i<N/2;i++)
    B[2*i]=A[N/2+1+i];
    }
    else{
    for(i=1;i<N/2;i++)
    B[2*i]=A[N/2+i];
    }
    
   for(i=0;i<N-1;i++){
       sum+=ab(B[i],B[i+1]);
   }
    
    printf("%d",sum);
    return(0);
}