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 m_1e9 = int_of_float 1e9 + 7
let ( +^ ) x y = (x + y) mod m_1e9
let ( -^ ) x y = x - y + if x < y then m_1e9 else 0
let ( *^ ) x y = ((x mod m_1e9) * (y mod m_1e9)) mod m_1e9
let split_string ?(pattern=" ") = Str.split @@ Str.regexp pattern

let n = Scanf.sscanf (read_line ()) "%d" @@ fun n -> n
let a = read_line () |> split_string |> List.map int_of_string

let rec loop ans arr = function
  | [] -> Printf.printf "%d\n" @@ ans
  | x :: xs -> 
    let t = Array.fold_left (fun s i -> s + if i = x then 1 else 0) 0 arr in
    if arr.(0) = x then arr.(0) <- arr.(0) + 1
    else if arr.(1) = x then arr.(1) <- arr.(1) + 1
    else arr.(2) <- arr.(2) + 1;
    loop (ans *^ t) arr xs

let () = loop 1 (Array.make 3 0) a