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.util.ArrayDeque;
import java.util.Deque;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner scan = new Scanner(System.in);

		long n = scan.nextLong();
		int a = scan.nextInt();
		int b = scan.nextInt();

		int countred = 0;
		int countblue = 0;

		Deque<String> colors = new ArrayDeque();

		if(a ==0 || b ==0 || n-a-b == 0) {
		for(int i=0;i<b;i++) {
			colors.addLast("red");
			countred++;
		}

		for(int i =0;i<a;i++) {
			colors.addLast("blue");
			countblue++;
		}
		}else if(n > a+b){

				for(int i =0;i<a;i++) {
					colors.addLast("blue");
					countblue++;
				}
				for(int i=0;i<b;i++) {
					colors.addLast("red");
					countred++;
				}
				for(int i = 0;i<n-a-b;i++) {
					colors.addLast("blue");
					countblue++;
				}
		}


		System.out.println(countblue);
		scan.close();
	}

}