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: Runtime Error)

let tuple3 a b c = (a, b, c)
;;

let tuple4 a b c d = (a, b, c, d)
;;

let rec f a_list min n m list =
  if Array.length a_list == n
  then
    list
    |> List.filter_map
      (fun (a, b, c, d) -> if a_list.(b-1) - a_list.(a-1) == c then Option.some d else Option.none) 
    |> List.fold_left (+) 0
  else
    List.init (m - min + 1) ((+) min)
    |> List.map (fun min -> f (Array.append a_list [|min|]) min n m list)
    |> List.fold_left max 0
;;

let () =
  let n, m, q = Scanf.scanf "%d %d %d\n" tuple3 in
  let list = List.init q (fun _ -> Scanf.scanf "%d %d %d %d\n" tuple4) in
  let res = f [||] 1 n m list in
  Printf.printf "%d\n" res
;;