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: Runtime Error)

C = {}

def f(n)
  return C[n] if C[n]

  return 1 if n <= 1

  C[n] = n * f(n - 1)
end

def calc(n)
  return 1 if n == 1

  a = 1
  v = 1
  while v * 2 <= n
    s = v + (n - v * 2)
    
    # sCv
    _a = f(s)
    _b = f(s - v)
    _c = f(v)

    a += (_a / _b) / _c

    #a += (f(s) / f(s - v)) / f(v)

    v += 1
  end
  a
end

n, m = gets.strip.split(" ").map(&:to_i)

as = []
before = -1
m.times do
  a = gets.to_i
  if before + 1 == a
    puts 0
    exit 0
  end
  before = a
  as << a
end

ans = 1
p = 0

while as.any?
  a = as.shift
  m = a - p - 1

  r = 
    if m == 0
      1
    else
      calc(m)
    end
    
  ans *= r

  p = a + 1
end

if p < n
  ans *= calc(n - p)
end

puts ans % 1000000007