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: Time Limit Exceeded)

N, M = [int(nm) for nm in input().split()]
A = [-1]
for m in range(M):
  A.append(int(input()))
A.append(N+1)

import math
total_case = 1
for m in range(M+1):
  a1 = A[m]
  a2 = A[m+1]
  safe_stairs = a2 - a1 - 1
  if safe_stairs == 0:
    print(0)
    break
  elif safe_stairs==1 or safe_stairs==2:
    total_case *= 1
    continue
  howmany_jump1 = safe_stairs - 1
  mx_jump2 = howmany_jump1 // 2
  #２のジャンプは０回～（mx_jump2）回
  case = 0
  for j2 in range(mx_jump2 + 1):
    jump2 = j2
    jump1 = howmany_jump1 - (j2 * 2)
    c = math.factorial(jump1 + jump2) // (math.factorial(jump1) * math.factorial(jump2))
    case += c
  total_case *= case
else:
  ans = total_case % (10**9 + 7)
  print(ans)