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: Accepted)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);

        // ##### 初期読み込み #######################################################################
        // 1行 文字 複数 ----------------------------------------------------------------------------
        String[] stringArray = scanner.nextLine().split(" ");

        long n = Long.parseLong(stringArray[0]);
        long a = Long.parseLong(stringArray[1]);
        long b = Long.parseLong(stringArray[2]);


        // ##### メイン処理 #######################################################################

        long amari = n%(a+b);
        long result = (n-amari)/(a+b)*a;

        if (amari > a) {
            result += a;
        } else {
            result += amari;
        }
        System.out.println(result);
    }
}