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: Wrong Answer)

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

public class Main {
  	public static void main(String[] args){
      //入力値の取得
      Scanner sc = new Scanner(System.in);
      Double n = Double.valueOf(sc.next());
      Double a = Double.valueOf(sc.next());
      Double b = Double.valueOf(sc.next());
      sc.close();

      if (a==0 && b==0) {
    	  System.out.println(0);
    	  return;
      }

      double answer = n/(a+b)*a;
      if (n%(a+b) <a) {
    	  answer+=n%(a+b);
      } else {
    	  answer+=a;
      }

      System.out.println(BigDecimal.valueOf(answer).setScale(0, RoundingMode.DOWN));
  	}
}