Problem:
Score : 200 points

Problem Statement

Takahashi has many red balls and blue balls. Now, he will place them in a row.

Initially, there is no ball placed.

Takahashi, who is very patient, will do the following operation 10^{100} times:

Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.

How many blue balls will be there among the first N balls in the row of balls made this way?

Constraints

1 \leq N \leq 10^{18}

A, B \geq 0

0 < A + B \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the number of blue balls that will be there among the first N balls in the row of balls.

Sample Input 1

8 3 4

Sample Output 1

4

Let b denote a blue ball, and r denote a red ball. The first eight balls in the row will be bbbrrrrb, among which there are four blue balls.

Sample Input 2

8 0 4

Sample Output 2

0

He placed only red balls from the beginning.

Sample Input 3

6 2 4

Sample Output 3

2

Among bbrrrr, there are two blue balls.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
    int blue, red;
    long long int N, amari, ans, num_full_set, nokori;
    
    scanf("%d %d %d", &N, &blue, &red);
    
    if(blue+red==0){
        printf("%d", 0);
        return 0;
    }
    
    amari = N%(blue+red);
    
    if (N>=(blue+red)) {
        amari = amari;
    }else{
        amari = 1+amari;
    }
    
    num_full_set = N/(blue+red);
    nokori = N-(num_full_set*(blue+red));
    
    if(amari == 0){
        ans = num_full_set*blue;
    }else{
        if(nokori>=blue){
            ans = blue+num_full_set*blue;
        }else{
            ans = num_full_set*blue+nokori;
        }
    }
    
    printf("%d", ans);
    
    return 0;
}