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 Rust (1.15.1) (judge verdict: Wrong Answer)

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    let mut it = buf.split_whitespace().map(|x| x.parse::<i64>().unwrap());
    let n = it.next().unwrap();
    let a = it.next().unwrap();
    let b = it.next().unwrap();
    let mut blue_cnt = 0;
    let mut all_cnt = 0;
    for _ in 0..100 {
        all_cnt += a;
        if all_cnt > n {
            blue_cnt += a - (all_cnt - n);
            break;
        } else {
            blue_cnt += a;
        }

        all_cnt += b;
        if all_cnt > n {
            break;
        }
    }
    println!("{}", blue_cnt);
}