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: Accepted)

N,M = map(int,input().split())
A = set([]); MOD = pow(10,9)+7
for i in range(M):
  a = int(input())
  A.add(a-1)
dp = [[0]*2 for _ in range(N)] #dp[i][j] i段目(0index)にj=0最後1段でj=1最後2段できた。
if 0 not in A:
  dp[0][0] = 1
if N == 1:
  print(sum(dp[0]))
  exit()
if 1 not in A:
  dp[1][0] = dp[0][0]+dp[0][1]
  dp[1][1] = 1
for i in range(2,N):
  if i in A:
    continue
  dp[i][0] = (dp[i-1][0]+dp[i-1][1])%MOD
  dp[i][1] = (dp[i-2][0]+dp[i-2][1])%MOD
ans = sum(dp[N-1])%MOD
#print(dp)
print(ans)