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 (7.3.0) (judge verdict: Runtime Error)

def m():
    MOD = 10 ** 9 + 7
    N, M = map(int, input().split())
    A = [int(input()) for _ in range(M)]
    dp = [0] * (N + 1)
    for a in A: dp[a] = -1
    if dp[1] != -1 and dp[2] != -1:
        dp[1] = 1
        dp[2] = 2
    elif dp[1] != -1 and dp[2] == -1:
        dp[1] = 1
    elif dp[1] == -1 and dp[2] != -1:
        dp[2] = 1
    else:
        return 0

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


print(m())