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: Runtime Error)

import java.util.Scanner;

class Main {
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		long N = Long.parseLong(sc.next());
		long A = Long.parseLong(sc.next());
		long B = Long.parseLong(sc.next());
//		String S = sc.next();

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

		StringBuilder sb = new StringBuilder();
		double aa = N <= Math.pow(10, 18) ? N: Math.pow(10, 18);
		boolean bb = true;
		int count = 0;

		for(int i=1; i <= aa;i++) {
			if(bb) {
				sb.append("b");
				count++;
			} else {
				count++;
			}
			if(bb && count == A) {
				bb =false;
				count = 0;
			} else if(!bb && count == B) {
				bb =true;
				count = 0;

			}
		}

		System.out.println(sb.toString().length());

	}
}