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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;


public class Main {


//	final int MOD = 1000000007;

	public static long ans = 0;

	public static void main(String[] args) {

//		int ans = 0;
		Scanner sc = new Scanner(System.in);
		boolean f = false;

		long n = sc.nextLong();
		long a = sc.nextLong();
		long b = sc.nextLong();

//		int n = sc.nextInt();


		//-------------------------------------------------------------------------------------

		ans = n / (a + b);
		ans *= a;

		if(n % (a + b) > 0) {
			if(n % (a + b) > a) {
				ans += a;
			} else
				ans += n % (a + b);
		}

		//-------------------------------------------------------------------------------------

//		if(f) {
//			System.out.println("Yes");
//		} else {
//			System.out.println("No");
//		}

		System.out.println(ans);

		sc.close();
	}
}