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: Accepted)

#include<bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
using namespace std;
typedef long long ll;
const int maxn=1e9;
void read(int &xxxxx){scanf("%d",&xxxxx);}
int a[100010];
ll sum[100010];

int main()
{
    int n;cin>>n;
    rep(i,1,n+1) read(a[i]);sort(a+1,a+n+1);
    rep(i,1,n+1) sum[i]=sum[i-1]+(ll)a[i];
    if(n%2==0) printf("%lld\n",2*sum[n]-4*sum[n/2]-(ll)(a[n/2+1]-a[n/2]));
    else{
        ll aa=2*sum[n]-4*sum[n/2]-(ll)a[n/2+1]-(ll)a[n/2+2];
        ll bb=2*sum[n]-4*sum[n/2+1]+(ll)a[n/2+1]+(ll)a[n/2];
        printf("%lld\n",max(aa,bb));
    }
    return 0;
}