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: Time Limit Exceeded)

import java.util.Scanner;

public class Main{

    public static void main(String[] args){
        // 変数宣言
        long total = 0;
        long blue = 0;
        long result = 0;
        long n;
        long a;
        long b;

        // 入力
        Scanner sc = new Scanner(System.in);
        
        n = Long.parseLong(sc.next());
        a = Long.parseLong(sc.next());
        b = Long.parseLong(sc.next());

        sc.close();


        //nよりa+bが小さい間だけ実行
        while( total < n ){
            for(long index=0; index < a ; index++){
                if( total < n){
                    blue++;
                    total++;
                }else{
                    break;
                }
            }
            for(long index=0; index < b ; index++ ){
                if( total < n){
                    total++;
                }else{
                    break;
                }
            }
        };

        // resultを出力
        result = blue;
        System.out.println( result );

    }
}