Problem:
Score : 300 points

Problem Statement

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order:

Choose two among A, B and C, then increase both by 1.

Choose one among A, B and C, then increase it by 2.

It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

Constraints

0 \leq A,B,C \leq 50

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the minimum number of operations required to make A, B and C all equal.

Sample Input 1

2 5 4

Sample Output 1

2

We can make A, B and C all equal by the following operations:

Increase A and C by 1. Now, A, B, C are 3, 5, 5, respectively.

Increase A by 2. Now, A, B, C are 5, 5, 5, respectively.

Sample Input 2

2 6 3

Sample Output 2

5

Sample Input 3

31 41 5

Sample Output 3

23

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

use std::cmp::max;
use std::fs::File;
use std::io::{stdout, BufWriter, Read, Write};
use std::thread;

fn main() {
    thread::Builder::new()
        .stack_size(32 * 1024 * 1024) // 32MB
        .spawn(run)
        .unwrap()
        .join()
        .unwrap();
}

/// ローカル環境なら _in.txt から入力を受ける
fn read_input() -> String {
    let stdin = std::io::stdin();
    let mut s = String::new();
    match option_env!("LOCAL") {
        Some(_) => File::open("src/_in.txt")
            .expect("File not found")
            .read_to_string(&mut s)
            .unwrap(),
        None => stdin.lock().read_to_string(&mut s).unwrap(),
    };
    s
}

fn run() {
    let out = stdout();
    let mut out = BufWriter::new(out.lock());
    let input = read_input();
    let mut input = input.split_whitespace();

    let in_a: i64 = input.next().unwrap().parse().unwrap();
    let in_b: i64 = input.next().unwrap().parse().unwrap();
    let in_c: i64 = input.next().unwrap().parse().unwrap();

    let rem = max(max(in_a, in_b), in_c) * 3 - in_a - in_b - in_c;
    if rem % 2 == 0 {
        writeln!(out, "{}", rem / 2).ok();
    } else {
        writeln!(out, "{}", rem / 2 + 2).ok();
    }
}