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)

def Fib(n):
    a, b = 0, 1
    if n == 1:
        return a
    elif n == 2:
        return b
    else:
        for i in range(n-2):
            a, b = b, a + b
        return b

a,b=[int(i) for i in input().split()]
ll=[]
y=-1
s=0
for i in range(b):
    x=int(input())
    ll.append(x)
    if x-y==1:
        s=1
        break
    y=x
ll.append(a+1)
l2=[]
if s==1:
    print(0)
elif b==0:
    print(Fib(a+2))
else:
    x=-1
    for i in ll:
        l2.append(Fib(i-x))
        x=i
    ans=1
    for i in l2:
        if i==0:
            i=1
        ans*=i
    print(ans%1000000007)