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: Accepted)

macro_rules! scan_sub {
    ( $itr:ident, $t:ty ) => {
        let mut buf = String::new();
        let $itr = {
            while buf.trim().len() == 0 {
                buf = String::new();
                std::io::stdin().read_line(&mut buf).unwrap();
            }
            buf.split_whitespace().map(|n| n.parse::<$t>().unwrap())
        };
    };
}

macro_rules! scan {
    ( $x:ident; Vec<$t:ty> ) => {
        scan_sub!(itr, $t);
        let $x = itr.collect::<Vec<_>>();
    };
    ( $( $x:ident ),*; $t:ty ) => {
        scan_sub!(itr, $t);
        let mut itr = itr;
        $(
            let $x = itr.next().unwrap();
        )*
    };
}

fn main() {
    scan!(n, a, b; usize);
    let mut n = n;

    let c = a + b;
    let mut ans = 0;
    if c < n {
        ans += a * (n / c);
        n = n % c;
    }

    ans += if a < n {a} else {n};

    println!("{}", ans);
}