Problem:
Score : 100 points

Problem Statement

You are given a positive integer N.
Find the minimum positive integer divisible by both 2 and N.

Constraints

1 \leq N \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum positive integer divisible by both 2 and N.

Sample Input 1

3

Sample Output 1

6

6 is divisible by both 2 and 3.
Also, there is no positive integer less than 6 that is divisible by both 2 and 3.
Thus, the answer is 6.

Sample Input 2

10

Sample Output 2

10

Sample Input 3

999999999

Sample Output 3

1999999998

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io;

/// 一行を読み込んで、空白で区切ったベクタにして返す
#[allow(dead_code)]
fn read<T>() -> Vec<T>
where
    T: std::str::FromStr,
    T::Err: std::fmt::Debug,
{
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    buf.trim()
        .split_whitespace()
        .map(|s| s.parse().unwrap())
        .collect()
}

/// n行を読み込んで、数字を空白で区切って二次元のベクタにして返す
#[allow(dead_code)]
fn read_table<T>(n: usize) -> Vec<Vec<T>>
where
    T: std::str::FromStr,
    T::Err: std::fmt::Debug,
{
    let mut ret: Vec<Vec<T>> = Vec::new();
    for _ in 0..n {
        ret.push(read::<T>());
    }
    ret
}

/// n行を読み込んで、1文字ずつを二次元のベクタにして返す
#[allow(dead_code)]
fn read_char_table(n: usize) -> Vec<Vec<char>> {
    let mut ret: Vec<Vec<char>> = Vec::new();
    for _ in 0..n {
        let mut buf = String::new();
        io::stdin().read_line(&mut buf).unwrap();
        let chr = buf.trim().chars().collect();
        ret.push(chr);
    }
    ret
}

/// n行を読み込んで、一次元のベクタにして返す
#[allow(dead_code)]
fn read_array<T>(n: usize) -> Vec<T>
where
    T: std::str::FromStr,
    T::Err: std::fmt::Debug,
{
    let mut array = Vec::new();

    for _ in 0..n {
        let mut buf = String::new();
        io::stdin().read_line(&mut buf).unwrap();
        array.push(buf.trim().parse().unwrap());
    }
    array
}

fn main() {
    // 数値
    let n = {
        let i = read::<isize>();
        (i[0])
    };

    let ans = if n % 2 == 0 { n } else { n * 2 };

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