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)

use std::io::*;

fn main() {
    let mut buf = String::new();

    stdin().read_line(&mut buf).unwrap();
    let n: u32 = buf.trim_right().parse().unwrap();

    buf.clear();
    stdin().read_line(&mut buf).unwrap();
    let mut ls: Vec<u32> = buf.clone()
                              .split_whitespace()
                              .map(|x| x.parse().unwrap())
                              .collect();
    ls.sort();

    let mut c: u32 = 0;
    for i in ls.iter().enumerate() {
        if i.0 % 2 == 0 {
            c += *i.1;
        }
    }

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