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=[]
b=0 in range(N)
for i in range(M):
    a.append(int(input()))
j=M-1
if j>=0 and a[0]==1:
    f1=0
    f2=1
elif j>=0 and a[0]==2:
    f1=1
    f2=1
else:
    f1=1
    f2=2
def f(n,a,j):
    if n>2:
        if j==-1 or ((n-1)!=a[j] and (n-2)!=a[j]):
            return (f(n-1,a,j)+f(n-2,a,j))%100000007
        if j>0 and a[j]==a[j-1]+1:
            return 0
        if j+1<M and a[j+1]==a[j]+1:
            return 0
        elif j>0 and (n-1)==a[j]:
            return f(n-2,a,j-1)%100000007
        elif (n-1)==a[j]:
            return f(n-2,a,j)
        elif (n-2)==a[j]:
            return f(n-1,a,j)%100000007
    elif n==1:
        return f1
    else:
        return f2
print(f(N,a,j)%100000007)