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: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long n = sc.nextLong();
		long a = sc.nextLong();
		long b = sc.nextLong();
		sc.close();
		// aがnより大きいパタン
		if (a > n) {
			System.out.println(a);
		}
		// aがnと等しいパタン
		if (a == n) {
			System.out.println(a);
		}
		// bがnより大きいばたん
		else if (b >= n) {
			System.out.println(a);
		}
		// a+bがnより以上のパタン
		else if ((a + b) >= n) {
			System.out.println(a);
		}
		// a+bがn以下のパタン
		else {
			long ab = a + b;
			long count = n / ab;
			long remainder = n % ab;
			long ans = a * count;
			if (remainder >= a) {
				ans = ans + a;
			} else {
				ans = ans + remainder;
			}
			System.out.println(ans);
		}
	}
}