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 Rust (1.15.1) (judge verdict: Accepted)

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
  let mut s = String::new();
  std::io::stdin().read_line(&mut s).ok();
  s.trim().split_whitespace()
    .map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
  let mut n = String::new();
  std::io::stdin().read_line(&mut n).ok();

  let mut ls = read_vec::<u32>();
  ls.sort();

  let ans: u32 = ls.iter()
    .enumerate()
    .filter(|&(i, _)| i % 2 == 0)
    .map(|(_, x)| x)
    .sum();

  println!("{}", ans);
}