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)

/*/ Author: kal013 /*/
// #pragma GCC optimize ("O3")
#include "bits/stdc++.h"
#include "ext/pb_ds/assoc_container.hpp"
#include "ext/pb_ds/tree_policy.hpp"
using namespace std;
using namespace __gnu_pbds;

template<class T> 
using ordered_set = tree<T, null_type,less<T>, rb_tree_tag,tree_order_statistics_node_update> ;

template<class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

struct custom_hash { // Credits: https://codeforces.com/blog/entry/62393
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
    size_t operator()(pair<int64_t,int64_t> Y) const{
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(Y.first * 31 + Y.second + FIXED_RANDOM);
    }
};

template<class T> ostream& operator<<(ostream &os, vector<T> V) {
    os << "[ ";
    for(auto v : V) os << v << " ";
    return os << "]";
}
template<class T> ostream& operator<<(ostream &os, set<T> S){
    os << "{ ";
    for(auto s:S) os<<s<<" ";
    return os<<"}";
}
template<class T> ostream& operator<<(ostream &os, unordered_set<T> S){
    os << "{ ";
    for(auto s:S) os<<s<<" ";
    return os<<"}";
}
template<class T> ostream& operator << (ostream& os, multiset<T> S){
    os << "{ ";
    for(auto s:S) os<<s<<" ";
    return os<<"}";
}
template<class T> ostream& operator<<(ostream &os, ordered_set<T> S){
    os << "{ ";
    for(auto s:S) os<<s<<" ";
    return os<<"}";
}
template<class L, class R> ostream& operator<<(ostream &os, pair<L,R> P) {
    return os << "(" << P.first << "," << P.second << ")";
}
template<class L, class R> ostream& operator<<(ostream &os, map<L,R> M) {
    os << "{ ";
    for(auto m:M) os<<"("<<m.first<<":"<<m.second<<") ";
    return os<<"}";
}
template<class L, class R, class chash = std::hash<R>> ostream& operator<<(ostream &os, unordered_map<L,R,chash> M) {
    os << "{ ";
    for(auto m:M) os<<"("<<m.first<<":"<<m.second<<") ";
    return os<<"}";
}
template<class L, class R, class chash = std::hash<R> > ostream& operator<<(ostream &os, gp_hash_table<L,R,chash> M) {
    os << "{ ";
    for(auto m:M) os<<"("<<m.first<<":"<<m.second<<") ";
    return os<<"}";
}

#define TRACE
#ifdef TRACE
    #define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){
        cerr << name << " : " << arg1 << endl;
    }
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){
        const char* comma = strchr(names + 1, ',');
        cerr.write(names, comma - names) << " : " << arg1<<" | ";
        __f(comma+1, args...);
    }
#else
    #define trace(...) 1
#endif

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

inline int64_t random_long(long long l,long long r){
    uniform_int_distribution<int64_t> generator(l,r);
    return generator(rng);
}
inline int64_t random_long(){
    uniform_int_distribution<int64_t> generator(LLONG_MIN,LLONG_MAX);
    return generator(rng);
}


/*/---------------------------Defines----------------------/*/
typedef vector<int> vi;
typedef pair<int,int> pii;

#define ll long long
#define F first
#define S second
#define pb push_back
#define endl "\n"
#define all(v) (v).begin(),(v).end()
/*/-----------------------Modular Arithmetic---------------/*/

const int mod=1e9+7;
inline int add(int x,int y){
    x+=y;
    if (x>=mod) return x-mod;
    return x;
}
inline int sub(int x,int y){
    x-=y;
    if (x<0) return x+mod;
    return x;
}
inline int mul(int x,int y){
    return (x*1ll*y)%mod;
}
inline int power(int x,int y){
    int ans=1;
    while(y){
        if (y&1) ans=mul(ans,x);
        x=mul(x,x);
        y>>=1;
    }
    return ans;
}
inline int inv(int x){
    return power(x,mod-2);
}
/*/-----------------------------Code begins----------------------------------/*/

void solve(){
    int n;
    cin >> n;
    vector<int> A(n);
    for(int i = 0; i < n; ++i) cin >> A[i];

    vector<pair<int,int> > B(n);

	for(int i = 0; i < n; ++i){
		B[i] = {A[i],i};
	}

	sort(all(B));

	reverse(all(B));
	set<int> idx;
	long long ans = 0;
	for(int i = 0; i < n; ++i){
		int z = B[i].second;
		idx.insert(z);

		if (idx.size() == 1){
			continue;
		}

		int l , r;
		auto it = idx.find(z);


		if (it != idx.begin()){
			--it;
			l = (*it);
		}
		else{
			l = (*idx.rbegin());

			assert(l > z);

		}

		it = idx.find(z);

		++it;
		if (it != idx.end()){
			r = (*it);
		}
		else{
			r = (*idx.begin());
			assert( r < z);
		}

		ans += min(A[l],A[r]);
	}
	cout << ans << endl;

}
int main(){
    // Use "set_name".max_load_factor(0.25);"set_name".reserve(512); with unordered set
    // Or use gp_hash_table<X,null_type>
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    cout<<fixed<<setprecision(25);
    cerr<<fixed<<setprecision(10);
    auto start = std::chrono::high_resolution_clock::now();
    int t=1;
    // cin>>t;
    while(t--) {
        solve();
    }
    auto stop = std::chrono::high_resolution_clock::now(); 
    auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start); 
    
    // cerr << "Time taken : " << ((long double)duration.count())/((long double) 1e9) <<"s "<< endl;     
}