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 PyPy3 (2.4.0) (judge verdict: Accepted)

###template###
import sys
def input(): return sys.stdin.readline().rstrip()
def mi(): return map(int, input().split())
###template###

N, M = mi()

#Nが1段なら、絶対に1通り
if N == 1:
  print(1)
  exit()

As = [int(input()) for _ in range(M)]
#0段目を含めて、階段を配列にする stairs[n]でアクセスする
stairs = ['.'] * (N+1)

#壊れた床を#に変え、1行の文字列にする
for a in As:
  stairs[a] = '#'
stairs = ''.join(stairs)

#そもそも無理なパターン。無理なパターンはこの場合だけ
if '##' in stairs:
  print(0)
  exit()


#n段目までの来方をDP[n]に格納していく
DP = [0]*(N+1)
#初期化
DP[0] = 1
#ループを楽にするため、DP[1]は求めておく ※1段しかない場合は排除済み
DP[1] = 1 if stairs[1] == '.' else 0

#Nに辿り着くまでループを回す
for i in range(2,N+1):
  DP[i] = (DP[i-2]+DP[i-1])%1000000007 if stairs[i] == '.' else 0

print(DP[N])