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<iomanip>
#include<cstdio>
#include<string>
#include<cstring>
#include<climits>
#include<algorithm>
#include<vector>
#include<cmath>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<deque>
#include<tuple>
#include<list>
#include<unordered_map>
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vii vector<int>
#define vll vector<ll>
#define lb lower_bound
#define pb push_back
#define mp make_pair
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep2(i,a,b) for(ll i=a;i<b;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define all(x) x.begin(),x.end()
#define INF (1 << 30) - 1
#define LLINF (1LL << 61) - 1
// #define int ll
using namespace std;
const int MOD = 1000000007;
const int MAX = 510000;


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,a[100010],ans=0;
    deque<ll> deq;
    cin>>n;
    rep(i,n){
        cin>>a[i];
    }
    sort(a,a+n);
    rep(i,n-1){
        if(i%4==0){
            deq.push_back(a[n-(i/2+1)]);
        }
        else if(i%4==1){
            deq.push_front(a[i/2]);
        }
        else if(i%4==2){
            deq.push_back(a[i/2]);
        }
        else{
            deq.push_front(a[n-(i/2+1)]);
        }
    }
    rep(i,n-2){
        ans+=abs(deq.at(i)-deq.at(i+1));
    }
    ans+=max(abs(deq.at(0)-a[n/2-(1-n%2)]),abs(deq.at(n-2)-a[n/2-(1-n%2)]));
    cout<<ans<<endl;
    return 0;
}