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(){
  int N,A,B;
  scanf("%d%d%d",&N,&A,&B);
  
  int putANum = 0;
  int junban = 0; //0でA(青)、1でB(赤)を置く番
  for(;;){
    if(junban == 0){
      if(N > A){
        putANum += A;
        N -= A;
        junban = 1;
      }else{ //A個全部は置けない場合、残り個数(N)を全部置く
      	putANum += N;
        break;
      }
    }else{
      if(N > B){
        N -= B;
        junban = 0;
      }else{
        break;
      }
    } 
  }
  printf("%d", putANum);
}