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.*;
import java.io.*;
import java.util.stream.*;

public class Main{
	private static BufferedReader in;
	private static BufferedWriter out;
	
	public static void main(String[] args) throws IOException {
		open();

		StringTokenizer st;
		StringBuilder sb;
		
		st=new StringTokenizer(readString());
		long N=Long.parseLong(st.nextToken());
		long A=Long.parseLong(st.nextToken());
		long B=Long.parseLong(st.nextToken());
		if(A==0)
			out.write(0+"\n");
		else if(B==0)
			out.write(A+"\n");
		else {			
			int monto=(int)(N/(A+B));
			int sobra=(int)(N-(monto*(A+B)));
			if(sobra>0)
				out.write(((A*monto)+sobra)+"\n");
			else
				out.write(((A*monto))+"\n");
		}
		close();
	}
	
	private static int[] readInts() throws IOException {
		return Stream.of(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
	}

	private static int readInt() throws IOException {
		return Integer.parseInt(in.readLine());
	}

	private static long[] readLongs() throws IOException {
		return Stream.of(in.readLine().split(" ")).mapToLong(Long::parseLong).toArray();
	}

	private static long readLong() throws IOException {
		return Long.parseLong(in.readLine());
	}

	private static double[] readDoubles() throws IOException {
		return Stream.of(in.readLine().split(" ")).mapToDouble(Double::parseDouble).toArray();
	}

	private static double readDouble() throws IOException {
		return Double.parseDouble(in.readLine());
	}

	private static String readString() throws IOException {
		return in.readLine();
	}

	private static int setBit(int S, int j) {
		return S | (1 << j);
	}

	private static int isOn(int S, int j) {
		return S & (1 << j);
	}

	private static int clearBit(int S, int j) {
		return S & ~(1 << j);
	}

	private static int toggleBit(int S, int j) {
		return S ^ (1 << j);
	}

	private static int lowBit(int S) {
		return S & (-S);
	}

	private static int setAll(int n) {
		return (1 << n) - 1;
	}

	private static int modulo(int S, int N) {
		return ((S) & (N - 1));
	} // returns S % N, where N is a power of 2

	private static boolean isPowerOfTwo(int S) {
		return (S & (S - 1)) == 0 ? true : false;
	}

	private static int nearestPowerOfTwo(int S) {
		return ((int) Math.pow(2.0, (int) ((Math.log((double) S) / Math.log(2.0)) + 0.5)));
	}

	private static int turnOffLastBit(int S) {
		return ((S) & (S - 1));
	}

	private static int turnOnLastZero(int S) {
		return ((S) | (S + 1));
	}

	private static int turnOffLastConsecutiveBits(int S) {
		return ((S) & (S + 1));
	}

	private static int turnOnLastConsecutiveZeroes(int S) {
		return ((S) | (S - 1));
	}

	private static void open() {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new BufferedWriter(new OutputStreamWriter((System.out)));
	}

	private static void close() throws IOException {
		out.flush();
		out.close();
		in.close();
	}
}