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: Time Limit Exceeded)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String[] S = sc.nextLine().split(" ");
        long N = Long.parseLong(S[0]);
        long A = Long.parseLong(S[1]);
        long B = Long.parseLong(S[2]);
        sc.close();

        long length = 0;
        long blueCount = 0;
        boolean isFinish = false;
        while(true){
            for(int i = 0; i < A; i++){
                length++;
                if(length > N){
                    isFinish = true;
                    break;
                }
                blueCount++;
            }

            for(int i = 0; i < B; i++){
                length++;
                if(length > N){
                    isFinish = true;
                    break;
                }
            }
            if(isFinish){
                break;
            }
        }

        System.out.println(blueCount);
                
    }

}