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: Wrong Answer)

#include <stdio.h>
#include <math.h>

int main(void) {
	double n;
	double a;//blue
	double b;//red
	double sum;
	double shou = 0;
	double amari = 0;
	scanf("%lf", &n);
	scanf("%lf", &a);
	scanf("%lf", &b);
	sum = a + b; //nの中にいくつあるか
	
	if(n >= sum) {
		shou = floor(n / sum);
		amari = floor(n - shou*sum);
	}
	else {
	if(n < a)
		amari = n;
	else
		amari = a;
	}
	if(amari >= a) {
		printf("%f\n", shou * a + a);
		return 0;
	}
	else {
		printf("%f\n", shou * a + amari);
		return 0;
	}
	return 0;
}