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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		long target = sc.nextLong();
		long b = sc.nextLong();
		long r = sc.nextLong();
		long sum = r + b;

		if (b == 0 || target == 0) {
			System.out.println(0);
			return;
		}

		//B+RがNよりも小さい場合
		if (target <= sum) {
			if(b <= target) {
				System.out.println(b);
				return;
			}else{
				System.out.println(target);
				return;
			}
		}

		//B+RがNよりも大きい場合
		long count = target / sum;
		long temp = target - (count * sum);

		if (b <= temp) {
			System.out.println(count * b + b);
		} else {
			System.out.println(count * b + temp);
		}

	}
}