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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner

fun main(args: Array<String>){
  val scn = Scanner(System.`in`) //Scannerのインスタンスを作成
  
  val n: Long = scn.nextLong()
  val a: Long = scn.nextLong()
  val b: Long = scn.nextLong()
  val g: Long = a + b
  var m: Long = n
  var result: Long = 0

  while(m > 0) {
    if (m >= g) {
      result += a
      m -= g
    } else if (m <= a) {
      result += m
      m -= m
    } else {
      result += a
      m -= m
    }
  }
  println(result)
}