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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a = sc.nextInt();
        int b = sc.nextInt();
        
        if (a == 0) {
            System.out.println(a);
            return;
        }
        if (n - (a + b) > 0) {
            // 数える数がボール投入数より多い場合
            int c = n / (a + b);
            int d = n % (a + b);
            if (d - a > 0) {
                System.out.println((c * a) + (a - d));
            } else {
                System.out.println((c * a) + d);
            }
        } else {
            // 数える数がボール投入数より少ない場合
            if (n - a > 0) {
                System.out.println(a);
            } else {
                System.out.println(n);
            }

        }
        sc.close();
    }
}