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 (1.3.71) (judge verdict: Accepted)

fun readString() = readLine()!!
fun readInt() = readLine()!!.toInt()
fun readLong() = readLine()!!.toLong()
fun readIntList() = readLine()!!.split(" ").map(String::toInt)
fun readLongList() = readLine()!!.split(" ").map(String::toLong)
fun <T> Array<Array<T>>.print() {
    this.forEach {
        println(it.joinToString(" "))
    }
}

fun main() {
    val (n, a, b) = readLongList()
    val period = a + b
    val res = (n / period) * a + (n % period).coerceAtMost(a)
    println(res)
}