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)

import java.io.PrintWriter
import java.util.*

val pw = PrintWriter(System.out)
val sc = Scanner(System.`in`)


fun main(args: Array<String>) {
    var N = sc.nextInt()
    val A = sc.nextInt()
    val B = sc.nextInt()

    var ans = 0

    while(N >= 0){
        if(N >= A){
            ans += A
            N -= A
        }else{
            ans += N
            N -= A
        }
        N -= B
    }

    println(ans)
}

fun func() {

}



fun next() = readLine()!!
fun nextInt() = next().toInt()
fun nextLong() = next().toLong()
fun nextDouble() = next().toDouble()
fun nextListOfString() = next().split(" ")
fun nextListOfInt() = nextListOfString().map { it.toInt() }
fun nextListOfLong() = nextListOfString().map { it.toLong() }
fun nextListOfDouble() = nextListOfString().map { it.toDouble() }