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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

Uses math;
var n, m, i, a : int64;
    dp : array[1..100010] of int64;
    remainder : int64;
begin
  remainder := 1000000007;
  readln(n, m);
  
  for i := 1 to 100010 do dp[i] := 0;
  for i := 1 to m do begin
    read(a);
    dp[a] := -1;
  end;
  dp[0] := 1;
  
  for i := 0 to 100005 do begin
    if dp[i] = -1 then Continue;
    if dp[i+1] <> -1 then dp[i+1] := (dp[i+1] + dp[i]) mod remainder;
    if dp[i+2] <> -1 then dp[i+2] := (dp[i+2] + dp[i]) mod remainder;
  end;
  writeln(dp[n]);
  for i := 0 to n do writeln(dp[i]);
end.