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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

var input = require('fs').readFileSync('/dev/stdin', 'utf8');
var input_arr = input.split(' ');
var n = parseInt(input_arr[0], 10);
var b = parseInt(input_arr[1], 10);
var r = parseInt(input_arr[2], 10);


var less = n % (b + r);

var sum = b * parseInt(n / (b + r), 10);
if (less <= b) {
sum += less;
} else {
sum += b;
}

console.log('%d', sum);