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>
int main()
{
  long N,A,B,i,ten,j,len;
  scanf("%ld%ld%ld",&N,&A,&B);
  ten=1;
  for(i=1;i<=50;i++)
  {
    ten=ten*10;
  }
  len=0;
  for(j=0;j<ten;j++)
  {
    if(j%2==0)
    {
      len=len+A;
      if(len>=N)
     {
       printf("%d",N-B*(j-1));
       exit(0);
     }
    }
    else
    {
      len=len+B;
      if(len>=N)
      {
        printf("%d",A+A*(j-1));
        exit(0);
      }
    }
   }
 }