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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.Scanner;
import java.util.Arrays;

public class Main {
    public static void main ( String[] args ) throws Exception {
        Scanner in = new Scanner( System.in );
        int N = Integer.parseInt( in.next() );
        int[] L = new int[ N ];
        for ( int i = 0; i < N; i++ ) {
            L[ i ] = Integer.parseInt( in.next() );
        }
        Arrays.sort( L );
        int cnt = 0;
        for ( int i = 0; i < N; i++ ) {
            if ( i % 2 == 0 ) {
                cnt += L[ i ];
            }
        }
        System.out.println( cnt );
    }
}