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)

/// Bismillahir Rahmanir Rahim
//Author: Tanvir Hussain
//ICE,NSTU
#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;
const long long MOD = 1000000007;
#define SET(x) memset(x, 0, sizeof(x))
#define SET2d(x,m,n) memset(x, 0, sizeof(x[0][0]) * m * n)
#define SETBOOL(x) memset(x,false,sizeof(x))
#define CLR(x) memset(x, -1, sizeof(x))
#define CLR2d(x,m,n) memset(x, -1, sizeof(x[0][0]) * m * n)
#define mp make_pair
#define PII pair<int, int>
#define pf printf
#define sf scanf
#define ALL(x) x.begin(),x.end()
#define pb push_back
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define np std::string::npos
#define highest(x) numeric_limits<x>::max()
#define lowest(x) numeric_limits<x>::min()
#define Inf INFINITY
#define minv(v) *min_element(v.begin(),v.end())
#define maxv(v) *max_element(v.begin(),v.end())
#define cases(cs,t) for(int cs=1;cs<=t;cs++)
#define PI acos(-1)
#define no1 __builtin_popcount
#define BOUNDARY(i, j) ((i >= 0 && i < row) && (j >= 0 && j < column))
#define uniq(vec) vec.resize(distance(vec.begin(),unique(vec.begin(),vec.end())))
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define sz(a) int(a.size())
#define ff first
#define ss second
#define endl "\n"
#define forch(it,s) for(auto it:s)
#define each(it,s) for(auto it = s.begin(); it != s.end(); ++it)
#define rep(i,a) for(int i=0; i<a;i++)
#define rep1(i,a,b) for(int i=(a);i<=(b);++i)
#define irep(i,b,a) for(int i=(b);i>=(a);--i)
#define bits(n) __builtin_popcount(n)
#define maxpq priority_queue<int>
#define minpq priority_queue<int, vector<int>, greater<int> >


int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a);}
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pair<int,int>> vpii;
typedef tree<pair<int, int>,null_type,less<pair<int, int>>,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1,-1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
const int maxx=100005;

//this fuction sorts vector pair according to first element in descending order.
bool sortinrev(const pair<int,int> &a,const pair<int,int> &b)
{
    return a.first>b.first;
}

template<typename T>inline T Bigmod(T base, T power, T MOD){
    T ret=1;
    while(power)
    {
        if(power & 1)ret=(ret*base)%MOD;
        base=(base*base)%MOD;
        power>>=1;
    }
    return ret;
}
double sq(double x) {return x*x;}
ll po(ll b,ll p){
    ll res=1;
    while(p){
        res*=b;
        p--;
    }
    return res;
}
ll lg2(ll x){
    ll res=0;
    while(x>1){
        res++;
        x/=2ll;
    }
    return res;
}
void solve(){
    int n;
    cin>>n;
    vll vec(n);
    rep(i,n) cin>>vec[i];
    sort(ALL(vec),greater<int>());
    ll ans=0;
    rep(i,n-1) ans+=vec[i];
    cout<<ans<<endl;
}

signed main()

{
    IOS;
    /*#ifndef ONLINE_JUDGE
        freopen ("input.txt","r",stdin);
        freopen ("output.txt","w",stdout);
    #endif*/
    int t;
    t=1;
    while(t--){
        solve();
    }

     return 0;

}
///Alhamdulillah