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: Runtime Error)

import java.io.IOException;
import java.util.Scanner;

public class Main {


	public static void main(String[] args) throws IOException {

		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int a = scan.nextInt();
		int b = scan.nextInt();
		int lineA = 0;
		int lineB = 0;
		while((lineA + lineB) < n) {
			lineA += a;
			lineB += b;
		}
		if((lineA + lineB) == n) {
			System.out.print(lineA);
		}else {
			lineA -=a;
			lineB -= b;
			int s = n - (lineA + lineB);
			if( a > s) {
				System.out.println(lineA + s);
			}else {
				System.out.println(lineA + a);
			}
		}

	}

}