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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <iostream>
using namespace std;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  long n,a,b;
  cin >>n>>a>>b;
  long balls = 0;
  long times = 0;
  while(balls <= n){
    balls += a+b;
    times++;
  }
  balls -= a+b;
  times--;

  if(n - balls >= a)
    cout << a*times + a<<endl;
  else
    cout << a*times + (n-balls) << endl;
  return 0;
}