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
from math import *
def fib(n):
    return round(( ((1 + sqrt(5))/2)**n - ((1 - sqrt(5))/2)**n ) / sqrt(5))
F = 10001
MD = 1000000007
ls = [0,1]
for f in range(F-2):
    s = sum(ls[-2:])
    ls.append(s)
def fib2(n):
    return ls[n]
old_a = 0
diff = []
N,M = map(int, input().split())
for m in range(M):
    a = int(input())
    if a-old_a==1 and old_a != 0:
        print (0)
        exit()
    diff.append(a-old_a)
    old_a = a
diff.append(N - old_a)
    
#print (diff)
ans = 1
for i,d in enumerate(diff):
    if i!=0 and i!=len(diff)-1:
        d-=1
    ans *= fib2(d)
    #ans %= 1000000007
print (ans % MD)
#print (ans)