Problem:
Score : 200 points

Problem Statement

An X-layered kagami mochi (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi.

Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

Constraints

1 ≤ N ≤ 100

1 ≤ d_i ≤ 100

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
d_1
:
d_N

Output

Print the maximum number of layers in a kagami mochi that can be made.

Sample Input 1

4
10
8
8
6

Sample Output 1

3

If we stack the mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, we have a 3-layered kagami mochi, which is the maximum number of layers.

Sample Input 2

3
15
15
15

Sample Output 2

1

When all the mochi have the same diameter, we can only have a 1-layered kagami mochi.

Sample Input 3

7
50
30
50
100
50
80
30

Sample Output 3

4

Solution in OCaml (4.02.3) (judge verdict: Accepted)

(* O(n) *)
let list_to_array n a xs =
  let res = Array.make n a in
  List.iteri (fun i x -> res.(i) <- x) xs;
  res

(* O(n * log(n)) (n = |xs|) *)
let unique_info n xs =
  if n <= 0 then
    [], 0
  else
    let sorted_xs = list_to_array n 0 xs in
    Array.sort compare sorted_xs;
    let last = sorted_xs.(n - 1) in
    let sofar, count, _ = Array.fold_right
      (fun x (sofar, count, prev) ->
        if x <> prev then
          x :: sofar, count + 1, x
        else
          sofar, count, x)
      sorted_xs
      ([last], 1, last)
    in
      sofar, count

(* O(n * log(n)) (n = |xs|) *)
let unique_count n xs = snd (unique_info n xs)

(* O(n * log(n)) *)
let solve n ds = unique_count n ds

let fold_int f a n =
  let rec fold f a i n =
    if i >= n then
      a
    else
      fold f (f a i) (i + 1) n
  in
    fold f a 0 n

let read_ints n =
  fold_int
    (fun sofar _ -> read_int () :: sofar)
    []
    n |> List.rev

let _ =
  let n = read_int () in
  let ds = read_ints n in
  solve n ds |> string_of_int |> print_endline