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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        List<Long> list = new ArrayList<>();
        for(int i = 0; i < 3; i++){
            Long num = sc.nextLong();
            list.add(num);
        }
       	long N = list.get(0);
        long A = list.get(1);
        long B = list.get(2);
        long C = 0;
          
        if(A == 0){
            System.out.println("0");
        }else if(N <= A){
            System.out.println(N);        
        }else if(N <= A + B){
            System.out.println(A);
        }else{
            while(N >= A+B){
                C++;
                N = N-(A+B);
            }
            System.out.println(C*A + N);
        }
    }
}