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){
   long N,A,B,r=0;
   scanf("%ld %ld %ld",&N,&B,&A);
  
  if(B==0){
    printf("0\n");
    exit(0);
  }
  for(int i=0; i<N;i++){
    if(N-r>B){
      r=r+B;
    }
      else{
        printf("%ld\n",B*i+(N-r));
        break;
      }
      if(N-r>=A){
        r=r+A;
      }
      else{
        printf("%ld\n",B*(i+1));
            break;  
      }
    }
    return 0;
    }