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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.Scanner;


public class Main {

	//残り数問が結果合わない
	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);

		String strN = sc.next();
		String strA = sc.next();
		String strB = sc.next();
		// 1 <= N <= 10^18
		long N = Long.parseLong(strN);

		//A >= 0
		long A = Long.parseLong(strA);

		//B >= 0
		long B = Long.parseLong(strB);

		boolean flg = false;

		if(A == 0) {
			System.out.println(0);
			return;
		}

		long roop = N / (A+B);
		long amari = N % (A+B);
		if(amari > A) {
			amari = A;
		}

		System.out.println(roop * A + amari);

	}

}