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 PyPy3 (2.4.0) (judge verdict: Runtime Error)

class Solution(object):
	def __init__(self):
		self.ans = 0
	def search(self, pos, dest):
		if(pos > dest):
			return
		if(pos == dest):
			self.ans += 1
			return
		if(pos - dest % 3 == 0):
			self.ans += (pos // 3) * 2
			return
		self.search(pos+1, dest)
		self.search(pos+2, dest)

N, M = map(int, input().split())
A = []

prev = 0

for i in range(M):
	curr = int(input())
	A.append(curr)
	#prev = curr

A.append(N+1)

#print(A)

ans = 1

prev = 0
for i in range(len(A)):
	s = Solution()
	if(i == 0):
		prev = 0
	else:
		prev = A[i-1] + 1
	#print(prev)
	s.search(prev,A[i]-1)
	con = s.ans
	#print(con)
	ans *= con % 1000000007

print(ans % 1000000007)