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 (2.7.1) (judge verdict: Accepted)

MOD = 1000000007
n, m = gets.split.map(&:to_i)
a = m.times.map{gets.to_i} if m > 0

spls = Array.new(0)
fib = Array.new(n+1)
fib[0] = fib[1] = 1
for i in 2..n
  fib[i] = fib[i-1] + fib[i-2]
end

flag = 0
if m > 0
  for i in 0..(m-1)
    if i == 0
      if a[i] == 1
        spls.push(1)
      else
        spls.push(a[i] - 1)
      end
    else
      if a[i] - a[i-1] == 1
        flag = 1
        break
      else
        spls.push(a[i] - a[i-1] - 2)
      end
    end
  end
  if n - a[i] > 1
    spls.push(n - a[i] - 1)
  end
else
  spls.push(n)
end

ans = 1
if flag > 0
  puts 0
else
  spls.each do |sp|
    ans = (ans*fib[sp]) % MOD
  end
  puts ans
end