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: Wrong Answer)

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

        buffer.trim().parse().unwrap()
    };

    let mut len: Vec<u8> = {
        let mut buffer = String::new();
        std::io::stdin().read_line(&mut buffer).unwrap();

        buffer.split_whitespace().map(|i| i.parse().unwrap()).collect()
    };

    len.sort();
    len.reverse();

    let mut sum = 0;
    for i in 0..n {
        sum += len[i * 2 + 1];
    }
    println!("{}", sum);
}