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: Accepted)

#include <stdio.h>

int main() {
    /*
    long long x = 1;
    
    for(int i = 0; i < 40; ++i) {
        x *= 2;
        printf("%d %lld\n", i+1, x);
    }
    return 0;
    */

    long long n, a, b;
    
    scanf("%lld%lld%lld", &n, &a, &b);
    
    long long total = 0, resposta = 0;
    
    long long ciclosCompletos = n / (a+b);
    resposta = ciclosCompletos * a;
    long long sobra = n % (a+b);
    
    if (sobra <= a) 
        resposta += sobra;
    else
        resposta += a;
    /*
    
    for(;;) {
        if (total + a <= n) {
            total += a;
            resposta += a;
        } else {
            resposta += n - total;
            break;
        }
        
        if (total + b <= n)
            total += b;
        else
            break;
    }
    */
    
    printf("%lld\n", resposta);
    

    return 0;
}