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++ (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;


int main() {
  
  
  int N ,A ,B;
  cin >> N >> A >> B;


int result, ans, amari;
int num = 10^100;
for (int i=0; i < num; i++){
  if (A==0){
    ans = 0;
  }
  else{
    result = N/(A+B);
    amari = N%(A+B);
    ans = amari+A*result;
  }
}
cout << ans << endl;
}