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.*;
import java.lang.*;

public class Main{
	public static void main(String[] args){
		Scanner scan = new Scanner(System.in);
				
		long N = scan.nextLong();
		long A = scan.nextLong();
		long B = scan.nextLong();
		long k = 0;

		if(1<=N && N<=Math.pow(10,18) && 0<=A && 0<=B && 0<A+B && A+B<=Math.pow(10,18)){
			if(A+B<N){
				k = N/(A+B);
				
				if(N-k*(A+B)>A){
					System.out.println( A*k + (N-k*(A+B)) - (N-(k*(A+B)+A)));
				}
				else{
					System.out.println( A*k + (N-k*(A+B)));
				}
				
			}
			else{
				System.out.println(A);
			}
		}
	}
}