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.chomp.split.map(&:to_i)
A = Array.new(N, 0)  # 0 means no trap, 1 means existence of trap
M.times do
  i = gets.chomp.to_i
  i -= 1  # 0-indexed
  A[i] = 1
end

MOD = 1000000007

# p N, M
# p a

def solve
  dp = Array.new(N, 0)  # 0 initialized
  dp[0] = 1 if A[0] == 0
  dp[1] = 1 + dp[0] if A[1] == 0
  if (N <= 2) 
    p dp[N-1]
    return
  end

  0.upto(N-3).each do |i|
    if A[i+2] == 0
      dp[i+2] = dp[i+1] + dp[i]
      dp[i+2] %= MOD
    end
  end
  
  p dp[N-1]
end

solve