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: Runtime Error)

#include <stdio.h>
#include <math.h>
main(){
  int N,A,B;
  scanf("%d %d %d",&N,&A,&B);
  int sum,k,mul,l,d,count;
  sum=A+B;
  k=pow(10,100);
  mul=sum*k;
  int a[mul+1];
  for(l=0;l<k;l++){
    for(d=l*sum;d<=l*sum+A-1;d++){
      a[d]=1;
    }
    for(d=l*sum+A;d<=l*sum+sum-1;d++){
      a[d]=0;
    }
  }
  for(l=0;l<N;l++){
    if(a[l]==1){
      count = count + 1;
    }
  }
  printf("%d",count);
}