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: Wrong Answer)

N_M = list(map(int, input().split()))
N = N_M[0]
M = N_M[1]
dis = []
for _ in range(M):
    dis.append(input())

dp = [0 for i in range(N+2)]
dp[0] = 1
if 1 in dis:
    dp[1] = 0
else :
    dp[1] = 1

for i in range(N):
    if i+2 in dis:
        dp[i+2]=0 
    else:
        if i in dis and i+1 in dis:
            dp[i+2]=0 
        else:
            if i in dis :
                dp[i+2]=dp[i+1]%1000000007
            elif i+1 in dis :
                dp[i+2]=dp[i]%1000000007
            else :
                dp[i+2]=dp[i]%1000000007+dp[i+1]%1000000007

print(dp[N])