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 sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import sympy as sym
input = sys.stdin.readline
 
# bisect_left(lists, 3)
# bisect_right(lists, 3)

# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

def main():
	N,M = map(int, input().split(" "))
	#N = int(input())
	#S = input()
	#S = list(S)
	
	a = []
	for _ in range(M):
		tmp = int(input())
		if len(a) != 0:
			if tmp-1 == a[-1]:
				print(0)
				sys.exit()
		a.append(tmp)

	if M == 0:
		print(fb(N+1))
		sys.exit()

	base = 0
	ans = 1
	for i, vol in enumerate(a):
		if base == 0:
			ans = ans * fb(vol)
			base = vol
		else:
			ans = (ans% 1000000007) * (fb(vol-base)% 1000000007)
			base = vol

	if N != base:
		ans = (ans% 1000000007) * (fb(N-base)% 1000000007)

	print(ans % 1000000007)


def fb(num):
    """num番目のフィボナッチ数を返す
    """
    a, b = 1, 0
    for _ in range(num):
        a, b = a + b, a
    return b

if __name__ == "__main__":
	main()