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 = map(int,input().split())
a = []
for i in range(m):
    A = int(input())
    a.append(A)
mod = 1000000007
dp = [0] * (n+1)
dp[0] = 1#0段目へ行く移動方法は1通りしかない
if 1 not in a:#もし階段がぶっ壊れてるリストに1段目が入っていないならば1段目へ行く移動方法は1通り、そうでないなら0通り
    dp[1] = 1
else:
    dp[1] = 0
for i in range(2,n+1):#2段目から調べていく
    if i in a:#現在のi段目がぶっ壊れリストに含まれてるなら何もせず飛ばす
        continue
    dp[i] = dp[i-1] + dp[i-2]#フィボナッチで処理
    dp[i] %= mod#modで割っていく
print(dp[n])#n+1番目を出力する