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 integer;
    double fract;
    double remand;
    int result=0;
    
    int N;
    int A,B;
    
    scanf("%d%d%d",&N,&A,&B);
    
    integer = (int)N/(A+B);
    fract = (double)N/(A+B) - integer;
    remand = fract*(A+B) +1 ;
  
    if( N%(A+B) != 0 )
    {
        if(A >= remand)
        {
            result = remand;
        }
        else
        {
            result = A;
        }
    }

    printf("%d", (integer*A)+result );
    return 0;
}