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 <iostream>
#include <bits/stdc++.h>
#include <algorithm>
#define lld long long int 
#define mod 1000000007
#define FIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
int main() 
{
    FIO
    lld n,a,b;
    cin >> n >> a >> b;
    lld sum = a+b;
    if(a == 0)
    {
        cout<<0<<endl;
    }
    else if(b == 0)
    {
        cout<<n<<endl;
    }
    else
    {
        if(sum > n)
        {
            cout<<min(a,n)<<endl;
        }
        else
        {
            cout<<((int(floor(n/sum)) * a) + min(n%sum,a))<<endl;
        }
    }
	return 0;
}