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<vector>
#include<string>
#include<algorithm>
#include<queue>
#include<deque>
#include<cmath>
#include<map>
#include<unordered_map>
#include<set>
#include<cstring>
#include<iomanip> //cout << fixed << setprecision(15) << x << endl;

using namespace std;
typedef long long ll;
const ll INF = 1e9 + 6;
const ll MOD = 1e9 + 7;
const ll LLINF = 1e18;
#define Pint pair<ll, ll>
#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rnr(i,a,b) for(int i=int(a);i>=int(b);i--)
#define rep(i,b) rng(i,0,b)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
/* -- template -- */

int main() {
    int N; cin >> N;
    deque<ll> A(N);
    rep(i, N) cin >> A[i];
    sort(all(A));
    deque<ll> dQ;
    dQ.pb(A[0]);A.pop_front();
    int cnt = 0;
    while(!A.empty()) {
        if(cnt % 4 < 2) {
            if(cnt % 2 == 0) {
                dQ.pb(A.back());
                A.pop_back();
            }else {
                dQ.push_front(A.back());
                A.pop_back();
            }
        }else {
            if(cnt % 2 == 0) {
                dQ.pb(A.front());
                A.pop_front();
            }else {
                dQ.push_front(A.front());
                A.pop_front();
            }
        }
        ++cnt;
    }
    ll ans = 0;
    rep(i, N - 1) {
        ans += abs(dQ[i] - dQ[i + 1]);
    }
    cout << ans << endl;
}