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: Time Limit Exceeded)

/* ????????? */

#include <stdio.h>

int main(void) {
  unsigned long  N, A, B, AandB;
  unsigned long  count = 0;
  unsigned long  i = 0;
  scanf("%lu%lu%lu",&N, &A, &B);
  
  AandB = A+B;
  /* ????????????????????????????? */
  if(N/10000000 >= AandB){
    while(N/(i+=10000000) >= (AandB));
  	i -= 10000000;
  }
  while(N/(++i) >= (AandB));
  
  N -= (AandB)*--i;
  count = A*i;
  
  if(N < A)
    count += N;
  else
    count += A;
  printf("%lu\n",count);
  
  return 0;
}