Problem:
Score : 200 points

Problem Statement

Snuke is having a barbeque party.

At the party, he will make N servings of Skewer Meal.

Example of a serving of Skewer Meal

He has a stock of 2N skewers, all of which will be used in Skewer Meal. The length of the i-th skewer is L_i.
Also, he has an infinite supply of ingredients.

To make a serving of Skewer Meal, he picks 2 skewers and threads ingredients onto those skewers.
Let the length of the shorter skewer be x, then the serving can hold the maximum of x ingredients.

What is the maximum total number of ingredients that his N servings of Skewer Meal can hold, if he uses the skewers optimally?

Constraints

1≦N≦100

1≦L_i≦100

For each i, L_i is an integer.

Input

The input is given from Standard Input in the following format:

N
L_1 L_2 ... L_{2N}

Output

Print the maximum total number of ingredients that Snuke's N servings of Skewer Meal can hold.

Sample Input 1

2
1 3 1 2

Sample Output 1

3

If he makes a serving using the first and third skewers, and another using the second and fourth skewers, each serving will hold 1 and 2 ingredients, for the total of 3.

Sample Input 2

5
100 1 2 3 14 15 58 58 58 29

Sample Output 2

135

Solution in C++ (Clang 3.8.0) (judge verdict: Wrong Answer)

//
//  main.cpp
//  comp
//
//  Created by Brenda Aurelie on 6/14/20.
//  Copyright © 2020 Brenda Aurelie. All rights reserved.
//

#include <iostream>
using namespace std;

int main() {
    int n,temp;
    cout << "Enter the number of skewer and the number of length: " << endl;
    cin >> n;
    int arr[2*n];
    for(int i=0;i<2*n;i++){
       
        cin >> arr[i];
    }
    for (int i=0;i<2*n;i++){
        for(int j=i+1;j<2*n;j++){
            if(arr[i]>arr[j]){
                temp=arr[i];
                arr[i]=arr[j];
                arr[j]=temp;
            }
        }
    }
    
    int sum=0;
    for(int i=0;i<n;i++){
        sum += arr[i*2];
    } 
    cout << sum << endl;
    
    return 0;
}