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 Ruby (2.3.3) (judge verdict: Accepted)

n, m = gets.strip.split.map(&:to_i)
a = readlines.map {|line| line.strip.to_i }.sort

MOD = 10**9+7
#p n
#p a
dp=[0]*(n+1)

a.each do |hall|
  dp[hall] = -1
end

dp[0]=1
n.times do |index|
  next if dp[index]==-1

  if (0..n)===(index+1) && dp[index+1]>=0
    dp[index+1]=(dp[index]+dp[index+1])%MOD
  end
  if (0..n)===(index+2) && dp[index+2]>=0
    dp[index+2]=(dp[index]+dp[index+2])%MOD
  end
end

 p dp[n]%MOD