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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
typedef pair<ll,ll> l_l;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
const int INF=1001001000;
const int mINF=-1001001000;
const ll LINF=1010010010010010000;
template<class T> inline bool chmin(T& a, T b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template<class T> inline bool chmax(T& a, T b) {
if (a < b) {
a = b;
return true;
}
return false;
}
//グリッド：（典型）dp,dfs,bfs,最短経路,その他
int main(){
    int n;cin >> n;
    vl a(n);
    rep(i,n) cin >> a[i];
    vl ra=a;
    sort(a.begin(),a.end());
    sort(ra.rbegin(),ra.rend());
    ll ans=0;
    vl num;
    int j=0;
    rep(i,n-1){
        if(i%2==0){
            num.push_back(ra[j]);
        }
        else{
            num.push_back(a[j]);
        }
        if(i%2==1) j++;
    }
    rep(i,n-2){
        ans+=abs(num[i+1]-num[i]);
    }
    //rep(i,n-1) cout << num[i] << endl;
    if(abs(a[j]-num[0])>abs(a[j]-num[n-2])){
        ans+=abs(a[j]-num[0]);
    }
    else ans+=abs(a[j]-num[n-2]);
    cout << ans << endl;
    return 0;
}