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: Runtime Error)

import sys
import math

from collections import defaultdict
from collections import deque
def load(vtype=int):
    return vtype(input().strip())
def load_list(seplator=" ", vtype=int):
    return [vtype(v) for v in input().strip().split(seplator)]
def exit():
    import sys
    sys.exit(0)
def perm_sub(li, used):
    if len(li) == len(used):
        return [deque()]
    k = []
    for i in range(len(li)):
        if i in used:
            continue
        used.add(i)
        sub_list = perm_sub(li, used)
        for sub in sub_list:
            sub.appendleft(li[i])
        k.extend(sub_list)
        used.discard(i)
    return k
def perm_li(li):
    return perm_sub(li, set())
def perm_n(n):
    return perm_sub(list(range(n)), set())
def join_i(li, sep=""):
    return sep.join([str(e) for e in li])
def li2n(li):
    n, base = 0, 1
    for i in range(len(li)-1, -1, -1):
        n += li[i] * base
        base *= 10
    return n
def sli2ili(li):
    return [int(s) for s in li] 
def prime_list(n):
    li = [i for i in range(2, n+1)]
    for i in range(len(li)):
        if li[i] >= int(math.sqrt(n)):
            break
        if li[i] == -1:
            continue
        for j in range(i+1, len(li)):
            if li[j] % li[i] == 0:
                li[j] = -1
    return [n for n in li if n != -1]
def gcd(a, b):
    a, b = max(a, b), min(a, b)
    while True:
        r = a % b
        if r == 0: return b
        a, b = b, r
def lcm(a, b):
    return int(a * b / gcd(a,b))
def all_subset(li):
    s = []
    n = len(li)
    for bit in range(0, (1<<(n+1))):
        ss = set()
        for i in range(0, n):
            if (bit & (1 << i)):
                ss.add(li[i])
        s.append(ss)
    return s
def factorial(n):
    if n == 1:
        return 1
    return n * factorial(n-1)
def mCn(m, n):
    def factorial_ntimes(m, n):
        r = 1
        while n:
            r *= m
            m -= 1
            n -= 1
        return r
    return int(factorial_ntimes(m, n) / factorial(n))


n, m = load_list()
A = []
for _ in range(m):
    A.append(load())
A = set(A)

dp = dict()
def f(now, goal, A):
    if now > goal:
        return 0
    if now == goal:
        return 1
    
    cnt = 0
    if now + 1 in A:
        pass
    elif now + 1 in dp:
        cnt += dp[now+1]
    else: 
        dp[now+1] = f(now+1, goal, A) 
        cnt += dp[now+1]

    if now + 2 in A:
        pass
    elif now + 2 in dp:
        cnt += dp[now+2]
    else:
        dp[now+2] = f(now+2, goal, A) 
        cnt += dp[now+2]
    return cnt
   
cnt = f(0, n, A)
print(cnt % 1000000007)