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: Wrong Answer)

#include<stdio.h>
#include <limits.h>
int main() {
 long N=0,A=0,B=0,shou=0,blue=0;
   scanf("%ld",&N);
   scanf("%ld",&A);
   scanf("%ld",&B);
  
  if(N<(A+B)){
    shou=1;
  }else if(N>=(A+B)){
  	shou=N/(A+B);
  }
  blue=shou*A;
  if((N!=0)&&((A+B)!=0)&&(N%(A+B)<A)){
    blue=blue+(N%(A+B));
  }else if((N!=0)&&(N%(A+B)>=A)&&((A+B)!=0)){
    blue=blue+A;
  }
  
  printf("%ld",blue);
  
  return 0;
}