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>
int main (){
  int n,a,b;
  scanf("%d%d%d", &n,&a,&b);

  if(a+b==n){
    printf("%d\n",a);
    return 0;
  } else if (a+b<n){
    if (a==0){
      printf("0\n");
      return 0;
    }
    else if(a != 0){
      if (b==0){
        printf("%d\n", a+(n-a));
        return 0;
      }
      for (int i = 0; i < n; i++) {
        if( (a+b)*i > n){
          if(n-(a+b*i)>=a)
            printf("%d\n", a*i-a);
          if(n-(a+b*i)<a)
            printf("%d\n", a*i-(n-(a+b*i)));
          return 0;
        }
        if( (a+b)*i == n){
          printf("%d\n", a*i);
          return 0;
        }
      }
    }
  }
}