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<stdlib.h>
#include<string.h>

#define rep(i,n) for(int i=0;i<n;i++)

int main()
{
	long i = 1;
	long a, b, n;
	long ans = 0;
	long r = 0;
	scanf("%ld%ld%ld", &n, &a, &b);
	if (n <= a) {
		printf("%d", n);
		return 0;
	}
	else if (n <= (a + b)) {
		printf("%d", a);
		return 0;
	}
	else {
		r = n % (a + b);
		ans = n / (a + b);
		if (r >= a) {
			printf("%d", (ans + 1)*a);
			return 0;
		}
		else {
			printf("%d", ans*a + r);
			return 0;
		}
	}
}