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.util.Scanner;

class Main{
	
	public static void main(String args[]){
      
	  Scanner scanner = new Scanner(System.in);
      String input = scanner.nextLine();
      
	  String nums[] = input.split(" ");
	  
	  System.out.println(nums[1]);
	  
	  long blue = Long.parseLong(nums[1]);
	  long red = Long.parseLong(nums[2]);
	  long sum = Long.parseLong(nums[0]);
	  
	  long count = sum / (blue + red);
	  long amari = sum % (blue + red);
	  
	  if(amari < blue){
	    System.out.println((count * blue)+amari);
	  }else{
	    System.out.println((count * blue)+amari);
	  }
	  
	}
}