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: Accepted)

import java.math.BigInteger
import java.util.*

private fun readLn() = readLine()!! // string line
private fun readInt() = readLn().toInt() // single int
private fun readLong() = readLn().toLong() // single long
private fun readStrings() = readLn().split(" ") // list of strings
private fun readInts() = readStrings().map(String::toInt) // list of ints
private fun readLongs() = readStrings().map(String::toLong) // list of longs

fun main(args: Array<String>) {
    val (n, a, b) = readLongs()
    if (a == 0L) {
        println(0)
    } else {
        val m = n % (a + b)
        val r = ((n - m) / (a + b)) * a + Math.min(a, m)
        println(r)
    }
}