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: Accepted)

MOD = 10 ** 9 + 7

def solve():
    N, M = map(int, input().split())
    A = [int(input()) for _ in range(M)]
    
    if N == 1 and M == 0:
        return 1
    elif N == 1 and M == 1 and A[0] == 0:
        return 0

    l = [1 for _ in range(N+1)]
    for a in A:
        l[a] = 0
        
    dp = [0 for i in range(N+1)]
    dp[1] = 0 if 1 in A else 1
    dp[2] = 0 if 2 in A else dp[1] + 1

    for i in range(3, N+1):
        if l[i] == 0:
            dp[i] = 0
        else:
            dp[i] = dp[i-1] + dp[i-2]
            dp[i] %= MOD

    return dp[N] % MOD

print(solve())