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 (1.42.0) (judge verdict: Time Limit Exceeded)

use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {n: usize, a: usize, b: usize};

    let mut sum_length = 0;
    let mut sum_a = 0;
    let mut sum_b = 0;
    let mut t = true;

    while sum_length < n {
        if t {
            sum_a += a;
            sum_length += a;
            t = false;
        } else {
            sum_b += b;
            sum_length += b;
            t = true;
        }
    }

    if t {
        println!("{}", sum_a);
    } else {
        println!("{}", sum_a - (sum_length - n));
    }
}