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 {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ

		Scanner scanner=new Scanner(System.in);
		long n = scanner.nextLong();
		long a = scanner.nextLong();  //blue
		long b = scanner.nextLong();  //red
		long count=0;
		long ans=0;
//		long m = n;
//
//		while(m > 0) {
//			m -= a+b;
//			count++;
//		}
//
//		if(n - (a+b)*(count-1) > a) {
//			ans += a;
//		}else {
//			ans += n - (a+b)*(count-1);
//		}
//		ans += a * (count-1);
//
//		System.out.println(ans);
//
//
		long m = n/(a+b);
		long l = n%(a+b);
		ans += a * m;
		if(l > a) {
			ans += a;
		}else {
			ans += l;
		}

		System.out.println(ans);

		scanner.close();

	}

}