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)

fn main() {
    let s = {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        s.trim_right().to_owned()
    };
    let (n, a, b) = {
        let mut ws = s.split_whitespace(); // 空白区切りの単語に分解する
        let n: i64 = ws.next().unwrap().parse().unwrap(); // イテレータから値を取り出して整数に
        let m: i64 = ws.next().unwrap().parse().unwrap();
        let o: i64 = ws.next().unwrap().parse().unwrap();
        (n, m, o)
    };

    let r = n % (a + b);
    println!("{}", a * ((n - r) / (a + b)) + std::cmp::min(a, r));
}

// nをa+bで割って、商q 余りrとすると、a*q+min(a, r)