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 (1.42.0) (judge verdict: Accepted)

// -*- coding:utf-8-unix -*-

use proconio::input;

fn main() {
    input! {
        mut abc: [u8; 3],
    }

    abc.sort();
    let mut ans = 0;
    if abc.iter().all(|&x| x % 2 == 0) || abc.iter().all(|&x| x % 2 == 1) {
        ans = (abc[2] - abc[0]) / 2 + (abc[2] - abc[1]) / 2;
    } else {
        if abc[0] % 2 == abc[1] % 2 {
            abc[0] += 1;
            abc[1] += 1;
        } else if abc[0] % 2 == abc[2] % 2 {
            abc[0] += 1;
            abc[2] += 1;
        } else {
            abc[1] += 1;
            abc[2] += 1;
        }
        ans = (abc[2] - abc[0]) / 2 + (abc[2] - abc[1]) / 2 + 1;
    }
    println!("{}", ans);
}