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)

# coding: utf-8
# Your code here!
n,m=map(int,input().split())
key=0
flag=False
a=[int(input()) for i in range(m)]
for i in range(m-1):#answer 0
    if a[i+1]-a[i]==1:
        flag=True

def fb(num):
    """num番目のフィボナッチ数を返す
    """
    a, b = 1, 0
    for _ in range(num+1):
        a, b = a + b, a
    return b

if m!=0:
    key=fb(a[0]-1)

if key==0: key=1

for i in range(m-1):
    key*=fb(a[i+1]-a[i]-2)

if m!=0:
    key*=fb(n-a[len(a)-1]-1)

if m!=0 and flag!=True:
    print(key%1000000007)
elif flag==True:
    print(0)
else: print(fb(n-1))