Problem:
Score: 500 points

Problem Statement

N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.

The person numbered i says:

"In front of me, exactly A_i people are wearing hats with the same color as mine."

Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.

Since the count can be enormous, compute it modulo 1000000007.

Constraints

1 \leq N \leq 100000

0 \leq A_i \leq N-1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_N

Output

Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.

Sample Input 1

6
0 1 2 3 4 5

Sample Output 1

3

We have three possible combinations, as follows:

Red, Red, Red, Red, Red, Red

Blue, Blue, Blue, Blue, Blue, Blue

Green, Green, Green, Green, Green, Green

Sample Input 2

3
0 0 0

Sample Output 2

6

Sample Input 3

54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17

Sample Output 3

115295190

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

let rec fix f x = f (fix f) x
let ( *@) a b = a * b mod 1000000007
let _ =
  Scanf.scanf "%d" @@ fun n ->
  let a = Array.init n (fun _ -> Scanf.scanf " %d" ((+) 0)) in
  let hats = [|0; 0; 0|] in
  (1, 0) |> fix (fun f (d, i) ->
    if i >= n then d else begin
      let c = Array.fold_left (fun c h -> if a.(i) = h then c+1 else c) 0 hats in
      0 |> fix (fun f j ->
        if j >= 3 then ()
        else if hats.(j) = a.(i) then hats.(j) <- hats.(j) + 1
        else f (j+1));
      f (c *@ d, i+1)
    end)
  |> Printf.printf "%d\n"