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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>#include <iostream>

#include <cmath>

unsigned long long parallelogram_len(unsigned long long A, unsigned long long B);

unsigned long long parallelogram_len(unsigned long long A, unsigned long long B) {
  unsigned long long a, b, y;
  if(A < B) {
    a = A;
    b = B;
  } else {
    a = B;
    b = A;
  }
  
  if(b%a == 0) {
    y = 2 * std::floor(b / a) * a - a;
  } else {
    y = 2 * std::floor(b / a) * a + parallelogram_len(a, b % a);    
  }
  return y;
}

int main(void) {
  unsigned long long N, X;
  std::cin >> N >> X;
   
  std::cout << N + parallelogram_len(X, N - X) << std::endl;
  return 0;
}