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 read<T: std::str::FromStr>() -> T {
	let mut s = String::new();
	std::io::stdin().read_line(&mut s).ok();
	s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
	read::<String>().split_whitespace()
		.map(|e| e.parse().ok().unwrap()).collect()
}

fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
	(0..n).map(|_| read_vec()).collect()
}

pub trait LexicalPermutation {
	/// Return `true` if the slice was permuted, `false` if it is already
	/// at the last ordered permutation.
	fn next_permutation(&mut self) -> bool;
}
impl<T> LexicalPermutation for [T] where T: Ord {
	/// Original author in Rust: Thomas Backman <serenity@exscape.org>
	fn next_permutation(&mut self) -> bool {
		// These cases only have 1 permutation each, so we can't do anything.
		if self.len() < 2 { return false; }

		// Step 1: Identify the longest, rightmost weakly decreasing part of the vector
		let mut i = self.len() - 1;
		while i > 0 && self[i-1] >= self[i] {
			i -= 1;
		}

		// If that is the entire vector, this is the last-ordered permutation.
		if i == 0 {
			return false;
		}

		// Step 2: Find the rightmost element larger than the pivot (i-1)
		let mut j = self.len() - 1;
		while j >= i && self[j] <= self[i-1]  {
			j -= 1;
		}

		// Step 3: Swap that element with the pivot
		self.swap(j, i-1);

		// Step 4: Reverse the (previously) weakly decreasing part
		self[i..].reverse();

		true
	}
}

fn main() {
	let (n, a, b) = {
		let tmp: Vec<u64> = read_vec();
		(tmp[0], tmp[1], tmp[2])
	};

	let loop_num = (n / (a+b)) + 1;
	let mut count = (loop_num-1) * a;
	let tmp = (n - (loop_num-1) * (a + b));
	if tmp >= a {
		count += a;
	} else {
		count += tmp;
	}
	println!("{}", count);
}