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)

#include <bits/stdc++.h>

// definitions
#define __Imperial_Lord__ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define _cps CLOCKS_PER_SEC
#define f first
#define s second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define rep(i, l, r) for (long long i = (l); i < (long long)(r); i++)
#define repb(i, r, l) for (long long i = (r); i > (long long)(l); i--)
#define lowb(a, x) ( lower_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define uppb(a, x) ( upper_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define sum(a)     ( accumulate ((a).begin(), (a).end(), 0ll))
#define high(x) __builtin_popcountll(x)
#define pb push_back
#define pf push_front
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define p2(a, b) cout << a << " " << b << endl
#define p3(a, b, c) cout << a << " " << b << " " << c <<endl
#define p4(a, b, c, d) cout << a << " " << b << " " << c << " " << d <<endl
#define exec cerr<<"Time taken : "<<(ld)(clock())/_cps<<"s"<<endl
#define mod1 998244353
#define mod 1000000007
#define pi 3.1415926535897932384

using namespace std;

// alias name for common data types
typedef vector<long long> vl;
typedef vector<long double> vld;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector< pair<long long, long long> > vp;
typedef vector<tuple<long long , long long , long long> > vtup;
typedef deque <long long> dql;
typedef deque <char> dqc;
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;

const ll INF=1e18;

template <class ...Args>
istream &read_args(Args &...args) {return (cin >> ... >> args);}
#define read(...)   __VA_ARGS__; read_args(__VA_ARGS__)

// cin and cout overloaded for vectors
template<typename T>    istream& operator>>(istream& is,  vector<T>& v)
{
    for (auto& i : v) 
        is >> i;        
    return is;
}
template<typename T>             
ostream& operator<<(ostream& os,  vector<T>& v)
{
    for (auto& i : v) 
        os << i << ' '; 
        return os;
}

ll none(ll i,ll j,ll n,ll m)
{
	if((i==0&&(j==0||j==m-1)) || (i==n-1&&(j==0||j==m-1))) return 2;
	else if(i==0 || j==0 || i==n-1 || j==m-1) return 3;
	else return 4;
}

// code beigns in main
int main()
{
	__Imperial_Lord__
    ll t=1;
    // cin >> t;
    ll lo;
    while(t--)
    {
    	ll read(n);
    	vl a(n);
    	cin >> a;
    	sort(all(a),greater<ll>());

    	// p1(a);
    	ll ans=a[0];
    	ll c=1,cnt=0;
    	rep(i,2,n)
    	{
    		// p1(a[c]);
    		if(cnt<2) ans+=a[c],cnt++;
    		if(cnt==2) cnt=0,c++;
    	}

    	p1(ans);
    }	
    exec;
}