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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int[] L = new int[2*N];
        int temp = 0, result = 0;
        for (int i = 0; i < 2*N; i++) {
            L[i] = scanner.nextInt();
        }
        for (int i = 0; i < L.length; i++) {
            for (int j = i; j < L.length; j++) {
                if (L[i] > L[j]) {
                    temp = L[i];
                    L[i] = L[j];
                    L[j] = temp;
                }
            }
        }
        for (int i = 0; i < L.length; i=i+2) {
            result += L[i];
        }
        System.out.println(result);
    }
}