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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.lang.reflect.Array;
import java.util.*;
import java.io.* ;
/*
NOTES:
-more than 10 digits (10^10), use long
-prefix sum of 1's and -1's to control flow of highlighting a region
-log(a)/log(b) = log base b of a
-create a position array if trying to see if elements show up next to each other sequentially
 i.e. elements 1,2,3 are within 3 spots of each other in an array
      arr[scannedInt - 1] = i; (position of i at p[i] in original)
- use a freq array when counting instances of an element occurring
- increment the counter first before simulating anything turn-based (two player game)

 */

public class Main {
    static long mod = (long)(Math.pow(10, 9) + 7);
    public static String alpha = "zabcdefghijklmnopqrstuvwxy";
    public static void main(String[] args) throws IOException{
        Scanner scan = new Scanner(System.in);
        long t = 1;
        //System.out.println(t);
        while (t-- > 0){
            int n = scan.nextInt();
            int[] arr = new int[n];
            for(int i = 0; i<n; i++){
                arr[i] = scan.nextInt();
            }
            Arrays.sort(arr);
            //System.out.println(Arrays.toString(arr));
            for(int i = 0; i<n/2; i++){
                int temp = arr[i];
                arr[i] = arr[n-i-1];
                arr[n-i-1] = temp;
                //System.out.println(Arrays.toString(arr));
            }
            //System.out.println(Arrays.toString(arr));
            long sum = 0;
            for(int i = 0; i<n-1; i++){
                sum += arr[i];
            }
            System.out.println(sum);
        }
        scan.close();

    }


    static boolean isPalindrome(String s){
        for(int i = 0; i<s.length()/2; i++){
            if(s.charAt(i) != s.charAt(s.length()-i-1)) return false;
        }
        return true;
    }

    static ArrayList factors(long n ){
        ArrayList<Long> facts = new ArrayList<>();
        for(int i = 1; i<=(int)Math.sqrt(n); i++){
            if(n%i == 0){
                if( i == Math.sqrt(n)) facts.add((long)i);
                else{
                    facts.add((long)i);
                    facts.add(n/i);
                }
            }
        }
        Collections.sort(facts);
        return facts;
    }

    static boolean isPrime (int n){
        if(factors(n).size() ==2) return true;
        else return false;
    }





    static long GCD(long a, long b) {
        // Everything divides 0
        if (a == 0) return b;
        if (b == 0) return a;

        // base case
        if (a == b) return a;

        // a is greater
        if (a > b) return GCD(a-b, b);
        return GCD(a, b-a);
    }
    static long LCM(long a, long b){
        return (a*b)/(GCD(a,b));
    }
    static String rev(String s){
        char[] arr = s.toCharArray();
        for(int i = 0; i<(s.length()+1)/2; i++){
            char temp = arr[i];
            arr[i] = arr[s.length()-i-1];
            arr[s.length()-i-1] = temp;
        }
        String fin = new String(arr);
        return fin;
    }
    static long pow(long a, long N) {
        if (N == 0) return 1;
        else if (N == 1) return a;
        else {
            long R = pow(a,N/2);
            if (N % 2 == 0) {
                return R*R;
            }
            else {
                return R*R*a;
            }
        }
    }

    static long powMod(long a, long N) {
        if (N == 0) return 1;
        else if (N == 1) return a % mod;
        else {
            long R = powMod(a,N/2) % mod;
            R *= R;
            R %= mod;
            if (N % 2 == 1) {
                R *= a;
                R %= mod;
            }
            return R % mod;
        }
    }

}