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>

using namespace std;

typedef long long ll;
typedef pair<int,int> pint;
typedef pair<ll,ll> pll;
typedef pair<int,pint> ppint;
typedef pair<ll,pll> ppll;
typedef vector<int> vint;
typedef vector<ll> vll;

const double pi=3.141592653589793;
const int INF10 = 1000000001;
const ll INF15 = 1e15 +1;
const long long INF18 = 1e18 + 1;
const int mod = 1000000007;
//const int mod = 998244353;
const double EPS=0.00001;

#define rep(i, n) for (int i = 0; i < (ll)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (ll)(n); i++)
#define rep2(i,start,end) for(int i=(ll)start;i<=(ll)end;i++)
#define vrep(i, n) for(int i=(ll)n-1;i>=0;i--)
#define vrep1(i, n) for(int i=(ll)n;i>0;i--)
#define all(n) n.begin(),n.end()
#define pb push_back
#define debug(x) cerr << #x <<": " << x << '\n'
#define arep(it,a) for(auto it : a )

struct edge{
    edge(int s,int e,long long c){
        to=e;from=s;cost=c;
    }
    edge(int i,long long c=1){
        edge(-1,i,c);
    }
    int from;
    int to;
    long cost;
};
typedef vector<edge> edges;

//bを何回足せばaを超えるか(O(a/b))
//a+b-1/bとすればよし

//2進数表示したときの最高桁(O(log n))
int bi_max(long n){
    int m = 0;
    for (m; (1 << m) <= n; m++);
    m = m - 1;
    return m;
}
//bi_eに二進数表示したやつを代入(O(log^2 n))
//bitset<N> a(n)でnの二進数表示が得られて、a[i]=0or1でi番目のfragが立ってるかわかる
//x^n mod m (nが負の時は０)(O(log n))
long myPow(long x, long n, long m=mod){
    if (n < 0)
        return 0;
    if (n == 0)
        return 1;
    if (n % 2 == 0)
        return myPow(x * x % m, n / 2, m);
    else
        return x * myPow(x, n - 1, m) % m;
}
// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
struct mint{
    ll x; // typedef long long ll;
    mint(ll x = 0) : x((x % mod + mod) % mod) {}
    mint operator-() const { return mint(-x); }
    mint &operator+=(const mint a)
    {
        if ((x += a.x) >= mod)
            x -= mod;
        return *this;
    }
    mint &operator-=(const mint a)
    {
        if ((x += mod - a.x) >= mod)
            x -= mod;
        return *this;
    }
    mint &operator*=(const mint a)
    {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const { return mint(*this) += a; }
    mint operator-(const mint a) const { return mint(*this) -= a; }
    mint operator*(const mint a) const { return mint(*this) *= a; }
    mint pow(ll t) const
    {
        if (!t)
            return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1)
            a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod - 2); }
    mint &operator/=(const mint a) { return *this *= a.inv(); }
    mint operator/(const mint a) const { return mint(*this) /= a; }
    bool operator!=(const int a) { return this->x!=a; }
     bool operator==(const int a) { return this->x==a; }
};
istream &operator>>(istream &is, const mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination{
    vector<mint> fact, ifact;
    combination(int n) : fact(n + 1), ifact(n + 1)
    {
        assert(n < mod);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i)
            fact[i] = fact[i - 1] * i;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i)
            ifact[i - 1] = ifact[i] * i;
    }
    mint operator()(int n, int k)
    {
        if (k < 0 || k > n)
            return 0;
        return fact[n] * ifact[k] * ifact[n - k];
    }
};
template<class T>
bool maxin (T &a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>
bool minin (T &a,T b){if(a>b){a=b;return 1;}return 0;}
template<class M,class N> 
constexpr common_type_t<M,N> gcd(M a,N b){
    a=abs(a);b=abs(b);
    if(a < b) return gcd(b, a);
    M r;
    while ((r=a%b)) {
        a = b;
        b = r;
    }
    return b;
}
template<class M,class N>
constexpr common_type_t<M,N> lcm(M a,N b){
    return a*b/gcd(a,b);
}

const int N_MAX=200005;
int n,a[N_MAX];

void Main(){
    int x=0,y=INF10,z=1;
    //入力
    cin>>n;
    rep(i,n)cin>>a[i];
    //処理
    ll ans=0;
    sort(a,a+n,greater<int>());
    priority_queue<int,vector<int>> que;
    que.push(a[0]);
    rep1(i,n-1){
        int now=que.top();que.pop();
        ans+=now;
        que.push(a[i]);
        que.push(a[i]);
    }
    //出力
    cout << ans <<endl;
}

int main(){
    cin.tie(nullptr);
    cout<<fixed<<setprecision(12);
    Main();
}