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: Accepted)

input = gets.chomp.split(" ")
n = input[0].to_i
m = input[1].to_i

div = 1000000007

a = []
for i in 1..m do
    a.push(gets.chomp.to_i)
end

p0 = [0, 1]
for i in 2..n do
    p0.push((p0[i-1] + p0[i-2])% div)
end

if m == 0
    p (p0[n]+p0[n-1])%div
else
a.uniq!
step = [a[0]]
for i in 1..a.size-1
    step.push(a[i]-a[i-1]-1)
end
step.push(n-a[a.size-1])


ans = 1

for i in 1..step.size do
    ans *= p0[step[i-1]]
    ans = ans % div
end

p ans
end