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: Time Limit Exceeded)

#!/usr/bin/env ruby
# require 'pry'

# dpテンプレ
# 最小化
# dp = Hash.new{ Float::INFINITY }
# dp[i] = [dp[i], dp[i-1] + hoge].min
# dp[0] = 0
# 最大化
# dp = Hash.new{ 0 }
# dp[i] = [dp[i], dp[i-1] + hoge].max
# dp[0] = 0

dp = Hash.new{ 0 }
dp[0] = 0
ary=[]
n,m = gets.split(' ').map(&:to_i)
m.times do |num|
  ary<<gets.to_i
end
ary.each { |i| dp[i] = 0 }

1.upto(n) do |i|
  next if dp.keys.include?(i)
  if i == 1
    dp[1] = 1
    next
  end
  if i == 2
    if dp[1] == 0
      dp[2] = 1
      next
    else
      dp[2] = 2
      next
    end
  end
  dp[i] = [dp[i], (dp[i-1] + dp[i-2])].max
end

# p dp
puts dp[n] % 1_000_000_007