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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;

class Program
{
    static int N;
    static int[] L;

    static void Main(string[] args)
    {
        Console.Write(BBQEasy());
    }

    private static string BBQEasy()
    {
        readData();

        return UseGuzai().ToString();
    }

    /// <summary>
    /// データを読み込む
    /// </summary>
    private static void readData()
    {
        string[] line = Console.ReadLine().Split(' ');

        N = int.Parse(line[0]);

        line = Console.ReadLine().Split(' ');

        L = new int[2*N];

        for (int i = 0; i < line.Length; i++)
        {
            L[i] = int.Parse(line[i]);
        }
    }

    private static int UseGuzai()
    {

        Array.Sort(L);

        int sum = 0;

        for (int i = 0; i < L.Length; i+=2)
        {
            sum += L[i];
        }

        return sum;
    }
}