Problem:
Score : 300 points

Problem Statement

There is a staircase with N steps. Takahashi is now standing at the foot of the stairs, that is, on the 0-th step.
He can climb up one or two steps at a time.

However, the treads of the a_1-th, a_2-th, a_3-th, \ldots, a_M-th steps are broken, so it is dangerous to set foot on those steps.

How many are there to climb up to the top step, that is, the N-th step, without setting foot on the broken steps?
Find the count modulo 1\ 000\ 000\ 007.

Constraints

1 \leq N \leq 10^5

0 \leq M \leq N-1

1 \leq a_1 < a_2 < ... < a_M \leq N-1

Input

Input is given from Standard Input in the following format:

N M
a_1
a_2
.
.
.
a_M

Output

Print the number of ways to climb up the stairs under the condition, modulo 1\ 000\ 000\ 007.

Sample Input 1

6 1
3

Sample Output 1

4

There are four ways to climb up the stairs, as follows:

0 \to 1 \to 2 \to 4 \to 5 \to 6

0 \to 1 \to 2 \to 4 \to 6

0 \to 2 \to 4 \to 5 \to 6

0 \to 2 \to 4 \to 6

Sample Input 2

10 2
4
5

Sample Output 2

0

There may be no way to climb up the stairs without setting foot on the broken steps.

Sample Input 3

100 5
1
23
45
67
89

Sample Output 3

608200469

Be sure to print the count modulo 1\ 000\ 000\ 007.

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

open List 
(* 1行読み込みする関数 *)
(* i_inp : 1行読み込んで整数に変換, f_inp, s_inpもある *)
let inp : unit -> string = fun () -> 
	input_line stdin
let s_inp = inp
let i_inp : unit -> int = fun () ->
	int_of_string (inp ())
let f_inp : unit -> float = fun () ->
	float_of_string (inp ())
 
(* 1行読み込んで、空白で分割する関数 *)
let inp_list : unit -> string list = fun () ->
	Str.split (Str.regexp " ") (inp ())
let s_inp_list : unit -> string list = inp_list
let i_inp_list : unit -> int list = fun () ->
	map (fun x -> int_of_string x) (inp_list ())
let f_inp_list : unit -> float list = fun () ->
  map (fun x -> float_of_string x) (inp_list ())

(* n行読み込む関数 *)
let rec inp_loop : (unit -> 'a) -> int -> 'a list = fun inp_fun n ->
  if n = 0 then []
  else
    let x = inp_fun () in
    x :: inp_loop inp_fun (n - 1)

      
(* 関数本体 *)
let () =
  let [n; m] = i_inp_list () in
  let brokens = Array.make (n + 1) false in
  let rec inp_loop m =
    if m = 0 then ()
    else
      let x = i_inp () in
      (brokens.(x) <- true;
       inp_loop (m - 1)) in
  let () = inp_loop m in
  let memo = Array.make (n + 1) None in
  let rec loop a =
    match memo.(a) with
      None ->
       if brokens.(a) then 0
       else
         if a + 2 <= n then
           let x = (loop (a + 1) + loop (a + 2)) mod 1000000007 in
           memo.(a) <- Some x;
           x
         else if a + 1 <= n then
           let x = (loop (a + 1)) mod 1000000007 in
           memo.(a) <- Some x;
          x
         else
           (memo.(a) <- Some 1;
            1)
    | Some x -> x
  in
  print_int (loop 0);
  print_newline ()