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<iostream>
using namespace std;

int main(){
    int n,a,b,counta=0,countsum=0;
    cin >> n >> a >> b;

    //while (n > countsum + a + b){
    //    counta += a;
    //    countsum += a + b;
    //}

    counta = n/(a+b)*a;
    countsum = n/(a+b)*(a+b);

    if (n <= countsum + a) cout << counta + n - countsum;
    else if (n <= countsum + a + b ) cout << counta + a;
    return 0;


}