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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for(long long int i = 0; i < n; i++)
#define print(n) std::cout << n << std::endl
#define _print(n) std::cout << n
using namespace std;
const int N = 1000000;

int main() {
    long long int n,a,b; cin >> n >> a >> b;

    int ans_e = n * (a / (a + b));
    long long int count_blue = 0;
    long long int sum = 0;


    if(n % (a + b) == 0)print(ans_e);
    else{
        while (sum < n){
            sum += a;
            if(sum < n){
                count_blue += a;
            }else{
                count_blue += n - (sum - a);
                break;
            }
            sum += b;
        }
        print(count_blue);
    }
}