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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"fmt"
)

func main() {
	var N, A, B uint64
	fmt.Scanf("%d %d %d", &N, &A, &B)
	var ans uint64
	switch {
	//もし青と赤の合計値がNより大きい場合
	case A+B > N:
		ans = A
		fmt.Println(ans)
	//もし合計個数から青と赤の合計があまりなく割り切れた場合
	case N%(A+B) == 0:
		ans = N / (A + B) * A
		fmt.Println(ans)
	//もし合計個数から青と赤の合計が割り切れない場合
	case N%(A+B) != 0:
		ans = N / (A + B) * A
		//もし余りの数がAの一度に並べる個数より小さい場合
		if N%(A+B) < A {
			ans += N % (A + B)
			fmt.Println(ans)
			//もし余りの数がAの一度に並べる個数より大きい場合
		} else if N%(A+B) > A {
			ans += A
			fmt.Println(ans)
		}
	}

}