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: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader() {
			br =  new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while(st==null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
					
				}
				catch(IOException e) {
					e.printStackTrace();
				}
				
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	public static void main(String[] args) {
		FastReader ob = new FastReader();
		long n=ob.nextLong(),a=ob.nextLong(),b=ob.nextLong();
		long q=n/(a+b),r=n%(a+b);
		long ans=(q*a);
		if(a<r) {
			ans+=a;
		}
		else if(a>=r && a!=0) {
			ans+=r;
		}
		System.out.println(ans);
		
	}

}