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: Time Limit Exceeded)

fn get_line() -> String {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();
    line.trim().to_string()
}

fn main() {
    let nums = get_line().split_whitespace()
        .map(|n| n.parse().unwrap())
        .collect::<Vec<usize>>();
    let (n, a, b) = (nums[0], nums[1], nums[2]);
    let mut s = String::new();

    let mut count = 0;

    while s.len() <= n {
        if count < a {
            s += "b";
        } else {
            s += "r"
        }

        if count == (a + b) {
            count = 1;
        } else {
            count += 1;
        }
    }

    let sum = s.chars().filter(|&c| c == 'b').fold(0, |sum, _| sum + 1);
    println!("{}", sum);
}