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: Runtime Error)

N,M=map(int,input().split())
a=[]
for i in range(0,M):
    a.append(int(input()))
ans=1
s=[-1]*(N+1)
s[0]=1
def dp(n):
    if n==0:
        return 1
    elif n==1:
        return 0 if 1 in a else 1
    if s[n-1]!=-1:
        c=s[n-1]
    elif n-1 in a:
        s[n-1]=0
        c=0
    else:
        c=dp(n-1)%1000000007
        s[n-1]=c
    if s[n-2]!=-1:
        b=s[n-2]
    elif n-2 in a:
        s[n-2]=0
        b=0
    else:
        b=dp(n-2)%1000000007
        s[n-2]=b
    return (c+b)%1000000007
print(dp(N)%1000000007)