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.math.BigDecimal;
import java.math.RoundingMode;
import java.util.Scanner;

/**
 * TITLE : Count Balls
 * URL : https://atcoder.jp/contests/abc158/tasks/abc158_b
 */
public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        String N = sc.next();
        String A = sc.next();
        String B = sc.next();
        BigDecimal NBC = new BigDecimal(N);
        BigDecimal ABC = new BigDecimal(A);
        BigDecimal BBC = new BigDecimal(B);

        BigDecimal answer = NBC.divide(ABC.add(BBC), 0, RoundingMode.DOWN).multiply(ABC);
        if (ABC.compareTo(NBC.remainder(ABC.add(BBC))) >= 0) {
            answer = answer.add(NBC.remainder(ABC.add(BBC)));
        } else {
            answer = answer.add(ABC);
        }
        System.out.println(answer.toString());
        sc.close();
    }
}