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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int descending_compare(const void *a, const void *b){
    if (*(int*)a > *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

int ascending_compare(const void *a, const void *b){
    if (*(int*)a < *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}


int lower_bound(int *a, int n, int key){
    int left, mid, right;
    left = 0, right = n;
    mid = (left + right)/2;
    while ((left+1 != mid || mid+1 != right) && mid != left){
        if (key > a[mid]){
            left = mid;
        }else{
            right = mid+1;
        }
        mid = (left + right)/2;
    }
    if (a[left] >= key)return left;
    if (a[mid] >= key)return mid;
    if (a[right] >= key)return right;
    return n;
}

//greatest common divisor
unsigned long  gcd(unsigned long x, unsigned long y){
    if (y == 0){ 
        return x;
    }else if (x > y){
        return gcd(y, x % y);
    }else{
        return gcd(x, y % x);
    }
}



long long factorial(int x){
    long long rtn = 1;
    int i;
    for (i = x; i > 1; i--){
        rtn = (rtn*i);
    }
    return rtn;
}

int main(void){
    int a, b;
    scanf("%d %d", &a, &b);
    printf("%d\n", (a+b)%24);

    return 0;
}