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 Ruby (2.3.3) (judge verdict: Runtime Error)

input1 = gets.chomp
n,m = input1.split.collect{ |item| item.to_i}
alist = []
ans = 1
maxdif = 0
m.times do |i|
  alist[i] =  gets.chomp.to_i
  before = i==0 ? 0 : alist[i-1]
  dif = alist[i] - before
  maxdif = dif if dif > maxdif
  if ((before == alist[i] -1) && (i != 0))
    ans = 0
  end
end
alist.push(n+1)

dif = alist[-1] - alist[-2]
maxdif = dif if dif > maxdif
fib = [1,1]
maxdif.times do |nn|
  fib[nn+2] = fib[nn+1] + fib[nn]
end


if ans == 0
  puts ans
else
  calclist = []
  start = 0
  alist.each do |ai|
    num = ai - start - 1
    calclist.push(fib[num])
    start = ai+1
  end
  ans = 1
  calclist.each { |pat| ans = ans * pat}
  puts ans%1000000007
end