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>
int main(void){
int r,b,n,i;
  scanf("%d%d%d",&n,&b,&r);
  if(b==0||n==b+r){
    printf("%d",b);
    return 0;
  }
  while(1){
    if(n>b+r){
    for(i=0;i<b;i++){
   b++;
      if(b==0||n==b+r) break;
    }
     for(i=0;i<r;i++){
  r++;
      if(n==b+r) break;
    }
    }if(n==b+r) break;}
      printf("%d",b);
      return 0;
    }