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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program main
 implicit none
 integer(8) :: a,b,n
  read *, n,a,b
   if (n<=a+b) then
    if (mod(n,a+b)==0) then
     write(*,'(i0)') b
    else if (mod(n,a+b)<=a) then
     write(*,'(i0)') 0
    else if (mod(n,a+b)>a) then
     write(*,'(i0)') mod(n,a+b)-a
    end if
   else if (n>a+b) then
    if (mod(n,a+b)==0) then
     write(*,'(i0)') b*n/(a+b)
    else if  (mod(n,a+b)<=a) then
     write(*,'(i0)') (n-mod(n,a+b))/(a+b)
    else if (mod(n,a+b)>a) then
     write(*,'(i0)') b*(n-mod(n,a+b))/(a+b)+mod(n,a+b)-a
    end if
   end if
  stop
end program main