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 C++ (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <list>
#include <math.h>

using namespace std;

int main() {
	int N, A, B, count = 0, counter = 0;
	double X = pow(10, 100);

	cin >> N;
	cin >> A;
	cin >> B;

	for (int i = 0; i < X; i++) {
		for (int b = 0; b < A; b++) {
			if (count < N) {
				count++;
				counter++;
			}
			else {
				break;
			}
		}
		count += B;
		if (count >= N) {
			break;
		}		
	}

	cout << counter << endl;

	return 0;
}