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>
#include <iostream>
#include <cmath>
#include <vector>
#include <string>
using namespace std;

int main()
{
  int I, J, K, L, M;
  double i, j, k;
  int result;
  bool check;
  long long N, A, B;
  cin >> N >> A >> B;
  if (N < A)
  {
    cout << N;
    return 0;
  }
  if (N < A + B)
  {
    cout << A;
    return 0;
  }
  if (N % (A + B) == 0)
  {
    cout << A * (N / (A + B));
    return 0;
  }
  else
  {
    if (N % (A + B) <= A)
    {
      cout << (N % (A + B)) + (A * N / (A + B));
      return 0;
    }
    else
    {
      cout << A + (A * N / (A + B));
      return 0;
    }
  }
}