Problem:
Score : 100 points

Problem Statement

Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.

In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".

The current time is A o'clock, and a contest will begin in exactly B hours.
When will the contest begin? Answer in 24-hour time.

Constraints

0 \leq A,B \leq 23

A and B are integers.

Input

The input is given from Standard Input in the following format:

A B

Output

Print the hour of the starting time of the contest in 24-hour time.

Sample Input 1

9 12

Sample Output 1

21

In this input, the current time is 9 o'clock, and 12 hours later it will be 21 o'clock in 24-hour time.

Sample Input 2

19 0

Sample Output 2

19

The contest has just started.

Sample Input 3

23 2

Sample Output 3

1

The contest will begin at 1 o'clock the next day.

Solution in Python (3.4.3) (judge verdict: Runtime Error)

fn main() {
    let a: i32 = read();
    let b: i32 = read();
    let ans = (a+b) % 24;
    println!("{}", ans);
}

// 以下関数

use std::io::*;
use std::str::FromStr;
pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

pub const MOD: u64 = 10e9 as u64 + 7;

pub fn is_prime(n: i32) -> bool {
    if n < 2 {
        return false;
    } else if n == 2 {
        return true;
    }
    let mut i = 2;
    while i * i < n {
        if n % i == 0 {
            return false;
        } else {
            i += 1;
            continue;
        }
    }
    true
}

pub fn lcm(mut n: i32, mut m: i32) -> i32 {
    n = if n > m { m } else { n };
    m = if n > m { n } else { m };
    let mut i = n;
    while i % m != 0 {
        i += n;
    }
    i
}

pub fn abs(n: i32) -> i32 {
    if n >= 0 {
        n
    } else {
        -n
    }
}

pub fn max(n: i32, m: i32) -> i32 {
    if n >= m {
        n
    } else {
        m
    }
}

pub fn min(n: i32, m: i32) -> i32 {
    if n <= m {
        n
    } else {
        m
    }
}