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: Wrong Answer)

/* aman36 - Aman Raj */
 
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;
 
#define fast_io ios :: sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define AfterDecimal(n) cout << fixed << setprecision(n);
#define pb push_back
#define ll long long
#define ld long double
#define all(a) a.begin(), a.end()
#define pll pair<ll, ll>
#define pii pair<int, int>
#define F first
#define S second
#define bits(x) __builtin_popcount(x)
#define fori(n)  for (ll i = 0; i < n; i++)
#define forj(n)  for (ll j = 0; j < n; j++)
#define fork(n)  for (ll k = 0; k < n; k++)
#define forlp(k,x,n)  for (ll k = x; k <= n; k++)
#define vll vector <ll>
#define mll map <ll, ll>
#define vpll vector <pll>
#define vi vector <int>
#define mem(x, v) memset(x, v, sizeof(x));
#define int ll
#define endl "\n"

// priority_queue<ll, vector<ll>, less<ll>>pq;     // Max Heap
// priority_queue<ll, vector<ll>, greater<ll>>pq;  // Min Heap
 
// template <typename T> using ordered_set =
//  tree<T, null_type, less<T>, rb_tree_tag, 
//  tree_order_statistics_node_update>;
 
#ifdef DEBUG
#define debug(...) __f(#__VA_ARGS__, __VA_ARGS__)
template<typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cout << "# "<< name << " = " << arg1 << '\n';
}
template<typename Arg1, typename... Args>
void __f(const char* names, Arg1 && arg1, Args &&... args){
        const char* comma = strchr(names + 1 ,',');
        cout << "# ";
        cout.write(names, comma - names)<< " = " << arg1 << " | ";
        __f(comma + 1, args...);
}
#else
#define debug(...) 36
#endif
 
ll power(ll x, ll y, ll p) { 
    ll res = 1;
    x = x % p;  
    while (y > 0) {
        if (y & 1) 
            res = (res * x) % p; 
        y = y >> 1;
        x = (x * x) % p; 
    }
    return res;
}

bool isPrime(int n) 
{
    if (n <= 1) return false; 
    if (n <= 3) return true; 
    if (n % 2 == 0 || n % 3 == 0) return false; 
    for (int i = 5; i * i <= n; i = i + 6) 
        if (n % i == 0 || n % (i + 2) == 0) return false; 
    return true; 
}
 
const ll INF = (ll)(8e18);
const ll MOD = 1e9 + 7;
const ll MOD2 = 998244353;
const int M = 3000+5;
const int N = 3e5+5;
const int maxn = 1e5+5;

auto cmp = [](pll a, pll b) {
    return (min(a.F,a.S)) > (min(b.F,b.S));
};

void solve() {
    int n;
    cin >> n;
    int a[n];
    fori(n) cin >> a[i];

    sort(a, a+n, greater<int>());

    int i=0;
    i++;
    i++;

    int ans = a[0];

    set <pll, decltype(cmp)> s(cmp);

    s.insert({a[0], a[1]});
    for(; i<n; i++) {
        pll p = *s.begin();

        s.erase(s.begin());
        s.insert({p.F, a[i]});
        s.insert({p.S, a[i]});

        ans += min(p.F, p.S);
    }

    cout << ans << endl;

}
 
signed main() {
    fast_io;
    // #ifndef ONLINE_JUDGE
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    // #endif
 
    int tc = 1;
    // cin >> tc;
 
    for(int i=1; i<=tc; i++) {
        // cout << "Case #" << i << ": " ;
        solve();
    }
 
    #ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
    #endif
 
    return 0;
}
 
// read the question correctly
// read code carefully
// corner cases // int vs ll // cin vs scanf // clear structures // maths
// check if max, min set to INT_MAX instead of INF
// check constraint