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>
using namespace std;

using ll = long long ;
int main()
{
    ll N, A, B;
    cin >> N >> A >> B;

    if(A == 0 && B == 0) 
    {
        cout << 0 << endl;
        return 0;
    }
    if(A == 0) 
    {
        cout << 0 << endl;
        return 0;
    }

    ll quotient = N /(A+B);
    ll rem = N % (A+B);


    ll result = A*quotient ;
    if(rem <= A)
    {
        result += rem;
    }
    else
    {
        result += rem;
    }
    
    cout << result << endl;


    return 0;
}