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.*

fun main(args: Array<String>) {
    val scn = Scanner(System.`in`)
    val N = scn.nextLong()
    val A = scn.nextLong()
    val B = scn.nextLong()

    var ans = 0L
    var total = 0L

    while(total<=N){
        if(total+A <= N) {
            ans+= A
            total += A
        }else{
            ans += N-total
            println(ans)
            return
        }

        if(total+B <=N){
            total += B
        }else{
            println(ans)
            return
        }
    }
}