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 Ruby (2.3.3) (judge verdict: Runtime Error)

# 入力
n,a,b = gets.chomp.split.map(&:to_i)

# もとになる組み合わせをつくる
blue = Array.new(a,"blue")
red = Array.new(b,"red")

ball = blue+red

#エラー避けるためにn個の配列にする
c = n/(a+b).to_f
c = c.ceil

# n個までの配列を作る
ball = ball*c

# こたえ
iro = ball[0..n-1]
puts iro.count("blue")