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: Accepted)

MOD = 10**9 + 7
n, m = map(int, input().split())

ban = [False] * (n + 1)#踏めない階段の配列、最初は全てOK

for i in range(m):
    a = int(input())
    ban[a] = True#壊れている段をTrueに変える

dp = [0] * (n+1)#DPテーブルは余裕が必要
dp[0] = 1#最初は１通り
for i in range(n):
    dp[i] %= MOD#MODで割っておく
    if(ban[i]):dp[i] = 0#壊れている段に行く方法は無い
    if i + 1 <= n:#上がれるのは１段、または２段
        dp[i+1] += dp[i]#配る配列
    if i + 2 <= n:
        dp[i+2] += dp[i]

print(dp[n] % MOD)#n段目に行くのは何通りか