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 sys
import numpy as np
import math

N, M = list(map(int, input().split()))
A = np.zeros(M)

for i in range(M):
    A[i] = int(input())


# a が連続しているとだめ
for i in range(M-1):
    if A[i] == A[i+1]-1:
        print(0)
        sys.exit()

# 穴同士の間隔を考えて，積を取る
d = np.zeros(M+1)

for i in range(M-1):
    d[i+1] = A[i+1] - A[i] - 2

d[0] = A[0] - 1
d[-1] = N - A[-1] - 1


p = np.zeros(M+1)

for i in range(M+1):

    n = int(d[i] // 2)

    for j in range(1,n+1):
        # 2の数 j個
        # 1の数 n1 = d - 2j
        n1 = d[i] - 2*j
        pp = math.factorial(n1+j)/(math.factorial(j) * math.factorial(n1))
        p[i] += pp

    p[i] += 1

mm = 1000000007
p = p % mm

print(int(np.prod(p) % mm))