Problem:
Score : 400 points

Problem Statement

Given are integers A, B, and N.

Find the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N.

Here floor(t) denotes the greatest integer not greater than the real number t.

Constraints

1 ≤ A ≤ 10^{6}

1 ≤ B ≤ 10^{12}

1 ≤ N ≤ 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B N

Output

Print the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N, as an integer.

Sample Input 1

5 7 4

Sample Output 1

2

When x=3, floor(Ax/B)-A×floor(x/B) = floor(15/7) - 5×floor(3/7) = 2. This is the maximum value possible.

Sample Input 2

11 10 9

Sample Output 2

9

Solution in OCaml (4.10.0) (judge verdict: Time Limit Exceeded)

let a, b, n = Scanf.scanf "%f %f %f\n" @@ fun a b c -> a, b, c
let rec f x m =
  let nm = (floor (a*.x/.b)) -. a*.(floor(x/.b)) in
  if x < 0. then m
  else if nm >= m then f (x-.1.) nm
  else f (x-.1.) m
let () = print_int (int_of_float (f n 0.))