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: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
        Scanner in = new Scanner(System.in);
        int N = in.nextInt(); // the number of cells on the Y axis
        int X = in.nextInt(); // the number of cells on the Y axis
        int L = 0;
        //場合分け
        if(N/2 == X && (N % 2 == 0)){
        	L = X *3;
        }else if(X < N-X){
        	L = N + X;
        	L += (X / (N-(2* X))) * 3 *(N-(2* X)) - (N-(2* X));
        }else if(N/2 > X ){
        	L = cal(N,X);
        }else if(N/2 < X){
        	L = cal(N,N -X);
        }

        System.out.println(L);
	}
	static int cal (int N,int X){
		int L = 0;
    	L = N + X;
		L += (N-X)/X * 2 -X * 2;
		return L;
	}

}