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(void)
{
    unsigned long n,a,b;
    scanf("%lu%lu%lu", &n,&a,&b);
    if(a==0){
        printf("0");
        return 0;
    }else if(b==0){
        printf("%lu",n);
        return 0;
    }else if(n>(a+b)){
        unsigned long tmp=n/(a+b);
        unsigned long blue=tmp*a;
        if(n-(a+b)>a){
            blue+=a;
            printf("%lu",blue);
            return 0;
        }else{
            blue+=n-(a+b);
            printf("%lu",blue);
            return 0;
        }
    }else if(n>=a){
        printf("%lu",a);
        return 0;
    }else if(n<a){
        printf("%lu",n);
        return 0;
    }
    return 0;
}