Problem:
Score: 400 points

Problem Statement

Quickly after finishing the tutorial of the online game ATChat, you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the friendliness of Player i is A_i.

The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere.

When each player, except the first one to arrive, arrives at the place, the player gets comfort equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0.

What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?

Constraints

All values in input are integers.

2 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \dots A_N

Output

Print the maximum total comfort the N players can get.

Sample Input 1

4
2 2 1 3

Sample Output 1

7

By arriving at the place in the order Player 4, 2, 1, 3, and cutting into the circle as shown in the figure, they can get the total comfort of 7.

They cannot get the total comfort greater than 7, so the answer is 7.

Sample Input 2

7
1 1 1 1 1 1 1

Sample Output 2

6

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//                   ********** YATIN KWATRA **********

// AC BAKSHI RABBA

#include <bits/stdc++.h>
#define ll                      long long 
#define ld                      long double
#define ull                     unsigned long long 
#define FIO                     ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define endl                    "\n"
#define fo(i,a,b)               for(int i = a; i<=b ; i++)
#define rfo(i,a,b)              for(int i = a; i>=b ; i--)
#define vii                     vector<int>
#define vll                     vector<ll>
#define pq                      priority_queue
#define uom                     unordered_map                
#define all(v)                  v.begin(),v.end()
#define mp                      make_pair
#define pb                      push_back
#define pob                     pop_back
#define ff                      first
#define ss                      second
#define pii                     pair<int,int> 
#define pll                     pair<long,long>
#define mll                     map<ll,ll>
#define mii                     map<int,int>
#define vvii                    vector<vii>
#define vvll                    vector<vll>
#define mod                     1000000007
#define pi                      3.1415926535897932384626433832795

using namespace std;

void INPUT() 
{  
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
}

/*
    
   -> Check For Overflows
   -> Out of Bounds
   -> Initialisations of global arrays and variables
   -> Its not Always Complex, Try to simplify
   -> write it down and try to solve
   -> Have Faith in urself and in ur approach
   -> If it seems totally out of reach, switch to next :)
   -> And Come Back later, if possible 
*/





void solve()
{ 
        ll n;
        cin >> n;
        
        vll v(n);
        
        fo(i,0,n-1) cin >> v[i];

        sort(all(v),greater<ll>());

        ll ans = 0LL;

        ans += v[0];

        ll left = n-2, val;
        int id;

        fo(i,1,n-1)
        {
          if(left>=2LL)
          {
            val = v[i]+v[i];
            ans += val;
            left--;
            left--;
          }
          else
          {
            id = i;
            break;
          } 
     }
     
     if(left>0) ans += v[id];

        cout << ans << endl;
       
}
int main() {
    
    FIO
    INPUT();
    
    int t;
    t = 1;
    //cin >> t;
    
    while(t--)
    {
       solve();
    }
    return 0;
}