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>

int main(void)
{
  int n, a, b, div, mod;
  div = 0;
  mod = 0;
  scanf("%d %d %d", &n, &a, &b);
  if (n < 1 || (a == 0 && b == 0) || a < 0 || b < 0)
  {
    return 0;
  }
  div = n / (a + b);
  mod = n % (a + b);
  if (mod == 0)
  {
    printf("%d", a * div);
  }
  else if (mod < a)
  {
    printf("%d", a * div + mod);
  }
  else
  {
    printf("%d", a * div + a);
  }
  return 0;
}