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)


def endFib(n, a1=1, a2=0):
    if (n < 1):
        return a1
    else:
        return endFib(n-1, a1 + a2, a1)


n, m = map(int, input().split())
o = 1
p = 0
ft = int(input())
if ft < 3:
    p = ft
elif ft == 3:
    o = 2
    p = ft
elif ft > 3:
    o = endFib(ft-1)
    p = ft

for i in range(m-1):
    t = int(input())
    if t-p == 1:
        print(0)
        exit()
    elif t-p-1 == 1:
        p = t
        pass
    elif t-1-p == 2:
        p = t
        pass
    elif t-1-p == 3:
        o = o*2
    elif t-1-p > 3:
        o = o*endFib(t-2-p)
        p = t
if n-p > 3:
    o = o * endFib(n-p-1)
elif n-p == 3:
    o = o*2
print(o % 1000000007)