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)

# coding: utf-8
import os
import time
import re
import math


def main():
    global N,M,A,mod,memo
    N,M=map(int,input().split())
    A=[int(input()) for  a in range(M)] #a (1段~N-1段)
    start=time.time();#
    memo=[-1]*(N+1)
    
    mod=10**9+7
    print(dp(0))
    #print("{}ms".format(math.floor(time.time()*1000-start*1000)))#

def dp(index):
    global N,M,A,mod,memo
    if(index>N):return 0
    if(index==N):return 1
    if(A.count(index)>0):return 0
    if memo[index]==-1:
        memo[index]=(dp(index+1)+dp(index+2))%mod
    return memo[index]

main()