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: Time Limit Exceeded)

def xgcd(a, b):
    if b == 0:
        g = a
        x = 1
        y = 0
        return g, x, y
    else:
        g, s, t = xgcd(b, a%b)
        x = t
        y = s - a//b*x
        return g, x, y

def invmod(a, p):
    _, x, _ = xgcd(a, p)
    if x < 0:
        x = x + p
    return x

def count_route(n):
  if n < 0:
    return 0
  res = 0
  for i in range(n//2+1):
    res += f[n - i] * inv[i] * inv[n - 2* i]
  return res%const
    
n, m = map(int, input().split())
const = 1000000007
#nまでの階乗をハッシュテーブルに格納する
f = [1 for i in range(n + 1)]
inv = [1 for i in range(n + 1)]
for i in range(1, n+1):
  f[i] = f[i - 1] * i
  inv[i] = invmod(f[i], const)
prev = 0
ans = 1
for i in range(m):
  m = int(input())
  ans = (ans * count_route(m - 1 - prev))%const
  prev = m + 1
  if ans == 0:
    break
ans = (ans * count_route(n - prev))%const
print(ans)