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)

import math

N , M = map(int,input().split())

a = [-1] * (M+2)

for i in range(1,M+1) :
    a[i] = int(input())

flag = True # 可能かどうか
for i in range(0,M+1) :
    if a[i] + 1 == a[i+1] :
        flag = False
        break

if flag :
    a[M+1] = N + 1

    ans = 1

    for i in range(M+1) :
        length = a[i+1] - a[i] - 2 # 上がる長さ
        max2step = math.floor(length / 2) # 2段の回数
        count = 1
        for j in range(1,max2step+1) :
            step2 = math.factorial(j)
            step1 = math.factorial(length-2*j)
            count += math.factorial(length - j) / step1 / step2
        ans *= count

    print(int(ans % 1000000007))
else :
    print(0)