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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <stdlib.h>

using namespace std;


int IntCompareA(const void* pElem1, const void* pElem2)
{
    return *(const int*)pElem1 - *(const int*)pElem2;
}


int main(){
	int n;
	cin >> n;
	int a[n];
	for(int i=0;i<n;i++)
	{
    	int x;
    	cin >> x;
    	if(cin.eof())
        	break;
		
    	a[i]=x;
	}
	
	qsort(a,n,sizeof(int),IntCompareA);
	/*for(int i=0;i<n;i++){
		cout << a[i] << endl;
	}*/
	
	int gr=0;
	int ls=0;
	int sm=0;
	int ans;
	for(int i=0;i<n;i++){
		sm += a[i];
	}
	
	if(n%2==0){
		for(int i=0;i<n/2;i++){
			ls += a[i];
		}
		gr = sm-ls;
		cout<< (gr*2-a[n/2]) - (ls*2-a[(n/2)-1]);
	}else{
		for(int i=0;i<(n-1)/2;i++){
			ls += a[i];
			gr += a[n-1-i];
		}
		int k,l;
		k=2*gr+a[(n-1)/2]-a[(n+1)/2]-2*ls;
		l=-(2*gr-(2*ls-a[n]-a[n-1]+2*a[(n-1)/2]));
		if(k>l){
			cout << k << endl;
		}else{
			cout <<l << endl;
		}
	}
	
	
	return 0;
}