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: Wrong Answer)

N, M = map(int, input().split())
lista = [int(input()) for _ in range(M)]

sum = 1 # パターン総数
current = 0 # 現在の段数

# 連続で階段が壊れていたら上がるのは不可能
for i in range(M - 1):
  if lista[i + 1] - lista[i] == 1: 
    sum = 0
    break

# 連続では階段が壊れていない
if sum == 1:
  for a in lista:
    cnt = 0
    diff = a -current - 1
    for i in range(diff + 1):
      for j in range(diff // 2 + 1):
        if (i + 2 * j) == diff:
          cnt += 1
    current = a + 1
    sum *= cnt
    
    # 最後の段
    if a == lista[M - 1]:
      cnt = 0
      diff = N - current
      for i in range(diff + 1):
          for j in range(diff // 2 + 1):
            if (i + 2 * j) == diff:
              cnt += 1
      sum *= cnt

print(sum % 1000000007)