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 (3.4.3) (judge verdict: Runtime Error)

def fib_memo(n):
    memo = [0]*(n+1)

    def _fib(n):
        if n <= 1:
            return n
        if memo[n] != 0:
            return memo[n]
        memo[n] = _fib(n-1) + _fib(n-2)
        return memo[n]

    return _fib(n)
  
N, M = map(int, input().split())
A = list(map(int, input().split()))
B = [0]*(M)
for i in range(M):
  diff = 0
  if i == 0:
    diff = A[i] - 1
    B[0] = fib_memo(diff + 1)
  else if i == (M - 1):
    diff = N - A[M-1] -1
    B[M-1] = fib_memo(diff + 1)
  else:
    diff = A[i] - A[i-1]
    B[i] = fib_memo(diff + 1)
Ans = reduce(lambda a, b: a*b, B)
print(Ans % 1000000007)