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 Python (3.8.2) (judge verdict: Runtime Error)

def fib_memo(x):
  memo=[0]*(x+1)
  def fib(x):
    if x==0 or x==1:
      return 1
    elif memo[x]!=0:
      return memo[x]
    else:
      memo[x]=fib(x-1)+fib(x-2)
      return memo[x]
  return fib(x)

n,m=[int(x) for x in input().split()]
a=[-1]
for i in range(m):
  p=int(input())
  a.append(p)
a.append(n+1)
ans=1
c=1
for i in range(m+1):
  if a[i+1]-a[i]-2<0:
    print(0)
    c=0
    break
  elif a[i+1]-a[i]-2==0:
    ans*=1
  else:
    ans*=((fib_memo(a[i+1]-a[i]-2))%1000000007)
    ans=ans%1000000007
if c==1:
  print(ans)