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>

using namespace std;

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define eee(a) (cerr << (#a) << ": " << (a) << endl);


using ll = long long;
using pint = pair<int, int>;
using mint = map<int, int>;

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

const char sp = ' ';
const char cmm = ',';
const int MOD = 1e9+7;
const int INF = 1e9;
const ll LINF = 1e18;

ll mod(ll a, ll b){return (a%b+b)%b;}
ll gcd(ll a, ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b){return a*b/gcd(a,b);}
void Yes(){cout << "Yes" << endl;}
void No(){cout << "No" << endl;}
void Judge(bool b){b?Yes():No();}
void YES(){cout << "YES" << endl;}
void NO(){cout << "NO" << endl;}
void JUDGE(bool b){b?YES():NO();}
ll powMod(ll b, ll e, ll m){ll r=1;while(e>0){if(e&1)r=(r%m)*(b%m)%m;b=(b%m)*(b%m)%m;e>>=1;}return r;}
double distance(ll x1, ll y1, ll x2, ll y2){return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));}

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T>
void ppp(T n){cout << n << endl;}
template<typename T1, typename T2> ostream& operator<<(ostream& s, const pair<T1, T2>& p) {return s << "(" << p.first << ", " << p.second << ")";}
template<typename T> ostream& operator<<(ostream& s, const vector<T>& v) {
    int len = v.size();
    s << '[';
    for (int i = 0; i < len; ++i) {
        s << v[i]; if (i < len-1) s << ", ";
    }
    s << ']';
    return s;
}

struct aaa{
    aaa(){
        cin.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(15);
    };
}fastio;

int main(){

    int n;
    cin >> n;

    vector<int> a(n), b(n);
    rep(i, n) cin >> a[i];

    b = a;
    sort(a.rbegin(), a.rend());
    sort(b.begin(),b.end());

    ll ans;

    if(n%2){
        ll tmp1 = 0ll, tmp2 = 0ll;
        for(int i = 0; i < (n-1)/2; ++i){
            tmp1 += 2*a[i];
            tmp2 -= 2*b[i];
        }
        tmp1 -= a[(n-1)/2]+a[(n-1)/2+1];
        tmp2 += b[(n-1)/2]+b[(n-1)/2+1];
        for(int i = (n-1)/2+2; i < n; ++i){
            tmp1 -= 2*a[i];
            tmp2 += 2*b[i];
        }
        ans = max(tmp1, tmp2);
    }
    else{
        ll tmp1 = 0ll, tmp2 = 0ll;
        for(int i = 0; i < (n-1)/2; ++i){
            tmp1 += 2*a[i];
            tmp2 -= 2*b[i];
        }
        tmp1 += a[(n-1)/2]-a[(n-1)/2+1];
        tmp2 += b[(n-1)/2]-b[(n-1)/2+1];
        for(int i = (n-1)/2+2; i < n; ++i){
            tmp1 -= 2*a[i];
            tmp2 += 2*b[i];
        }
        ans = max(tmp1, tmp2);
    }

    ppp(ans);


    return 0;
}