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: Runtime Error)

fun main(args: Array<String>) {
    val inLine = readLine()?.split(" ").orEmpty()

    val n = inLine[0].toInt()
    val a = inLine[1].toInt()
    val b = inLine[2].toInt()
    var ball = ""
    var ballLine = ""
    var i = 0
    while(i < a){
        ball += "b"
        i++
    }
    i = 0
    while(i < b){
        ball += "r"
        i++
    }

    var count = 1
    i = 0
    run loop@{
        while (i < 100){
            if(count * (a + b) > n){
                return@loop
            }
            count *= 10
            i++
        }
    }
    i = 0
    while(i < count){
        ballLine += ball
        i++
    }

    val result = ballLine.dropLast(ballLine.length - n).replace("r","")
    println(result.length)


}