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: Runtime Error)

#include "bits/stdc++.h"
using namespace std;

int main() {
  int N, A, B;
  cin >> N >> A >> B;
  vector<char> balls;
  int b, r;
  for (int i = 0; i < true; i++) {
    for (int i = 0; i < A; i++) {
      balls.push_back('b');
      for (int i = 0; i < B; i++) {
        balls.push_back('r');
      }
    }
  }

  int count = 0;
  for (int i = 0; i < N; i++) {
    if (balls.at(i) == 'b') {
      count++;
    }
  }
  cout << count << endl;
}