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: Runtime Error)

import java.io.*;
import java.util.*;
public class Main {
  public static void main(String[] args) throws Exception {
    String line = null;
    long sum = 0L;
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    line = br.readLine();
    String data[] = line.split(" ");
    long n = Long.parseLong(data[0]);
    long x = Long.parseLong(data[1]);
    //System.out.println("N=" + n + ",X=" + x);
    long a = 0L; // 短辺の長さ
    long b = 0L; // 長辺の長さ
    long c = 0L; // 正三角形の数
    long d = 0L; // 長辺の残り
    long cnt = 0L;

    while (true) {
      if (cnt == 0L) {
        if (x >= (n - x)) {
          a = n - x;
          b = x;
        } else {
          a = x;
          b = n - x;
        }
      } else {
        b = a;
        a = d;
      }
      c = b / a;
      d = b % a;
      sum = sum + (c * 3L * a);
      //System.out.println("a=" + a + ",b=" + b + ",c=" + c + ",d=" + d + ",sum=" + sum);
      if (d == 0L) {
        break;
      }
      cnt = cnt + 1L;
    }
    System.out.println(sum);
  }
}