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 (3.8.2) (judge verdict: Runtime Error)

n, m=map(int, input().split())

from itertools import product

def mat2_mul(X, Y):
    Z = [ [0, 0], 
          [0, 0] ]
    for (i,j,k) in product(range(2),range(2),range(2)):
        Z[i][j] += X[i][k] * Y[k][j]
    return Z

def mat2_pow(X, n):
    if n == 0:
        return [ [1, 0],
                 [0, 1] ]
    elif n % 2:
        return mat2_mul(X, mat2_pow(X, n-1))  
    else:
        half_pow = mat2_pow(X, n/2)
        return mat2_mul(half_pow, half_pow)

def fb(n):
    if n == 0:
        return 0
    else:
        F = [ [0, 1],
              [1, 1] ]
        return mat2_pow(F, n-1)[1][1]
l=[]
for _ in range(m):  
  l.append(int(input()))
ans=fb(l[0])
#print(ans)
for i in range(m-1):
  if l[i]+1==l[i+1]:
    print(0)
    exit()
  elif l[i+1]-l[i]>2:
    ans=ans*(fb(l[i+1]-l[i]-1))
print(ans*fb(n-l[m-1])%1000000007)