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.split.map(&:to_i)
A = Hash.new {false}
M.times do
  A[gets.to_i] = true
end

MOD = 10 ** 9 + 7

dp = Array.new(N + 1) {0}
dp[0] = 1

for i in 0..N
  next_stair_1 = i + 1
  next if next_stair_1 > N

  if A[next_stair_1] == false
    dp[next_stair_1] += dp[i]
    dp[next_stair_1] %= MOD
  end

  next_stair_2 = i + 2
  next if next_stair_2 > N

  if A[next_stair_2] == false
    dp[next_stair_2] += dp[i]
    dp[next_stair_2] %= MOD
  end
end

puts dp[-1]