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)

N, M = list(map(int, input().split()))
a = []
for i in range(0, M):
    a.append(int(input()))
a.sort()
T = 10 ** 9 + 7

r = [0 for _ in range(N + 1)]
r[0] = 1
r[1] = 1


def fib(n):
    if r[n] > 0:
        return r[n]
    else:
        r[n] = fib(n - 1) + fib(n - 2)
        return r[n]


for i in range(0, M - 1):
    if a[i] + 1 == a[i + 1]:
        print(0)
        break
else:
    now = 0
    c = 1
    for i in range(0, M):
        c *= fib(a[i] - now - 1) % T
        c %= T
        now = a[i] + 1
    c *= fib(N - now)
    print(c % T)