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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#pragma warning (disable:4996)
#include <stdio.h>


int main()
{
	unsigned long long n;
	unsigned long long  a, b;
	scanf("%llu %llu %llu", &n, &a, &b);
	unsigned long long j = 0;
	bool c = true;
	unsigned long long tmp;
	unsigned long long d = 0;

	unsigned long long i = 0;
	do
	{
		j++;
		if (c)
		{
		blue:
			if (j >= a)
			{
				c = false;
				j = 0;
				goto red;
			}
			d++;
		}
		else
		{
		red:
			if (j >= b)
			{
				c = true;
				j = 0;
				goto blue;
			}
			j += b;
			i += b - 1;
		}
			i++;
	} while (i != n);
	printf("%lld", d);

	return 0;
}