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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

int main()
{
	long n, a, b;
	scanf("%ld%ld%ld",&n,&a,&b);
	long sum = 0;
	long mod;
	if (a == 0)
	{
		printf("0");
		return 0;
	}
	if (n >= a + b)
		mod = n % (a + b);
	while (n > 0)
	{
		if (n > a)
		{
			n -= a;
			sum += a;
		}
		else if (n == a)
		{
			n -= a;
			sum += a;
			break;
		}
		else if (n < a)
		{
			sum += n;
			break;
		}
		if (n > b)
			n -= b;
		else if (n == b)
		{
			n -= b;
			break;
		}
		else if (n < b)
		{
			n = 0;;
			break;
		}
	}
	printf("%ld", sum);
	return (0);
}