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)

let () =
    let main () =
        let n, m = Scanf.sscanf (read_line ()) "%d %d" (fun n m -> n, m) in
        let a = Array.init m (fun _ -> int_of_string (read_line ())) in
        let f = Array.make (n + 1) true in
        let () = Array.iter (fun v -> f.(v) <- false) a in
        let c = Array.make (n + 1) 0 in
        let () = c.(0) <- 1 in
        let mm = 1_000_000_007 in
        for i = 0 to n - 1 do
            if               f.(i + 1) then c.(i + 1) <- (c.(i + 1) + c.(i)) mod mm;
            if i + 2 <= n && f.(i + 2) then c.(i + 2) <- (c.(i + 2) + c.(i)) mod mm
        done;
        Printf.printf "%d\n" c.(n)
    in
    main ()