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 Rust (1.15.1) (judge verdict: Accepted)

fn main() {
    let (n, m) = {
        let mut line = String::new();
        std::io::stdin().read_line(&mut line).ok();
        let mut iter = line.trim().split_whitespace().map(|e| e.parse::<usize>().unwrap());
        (iter.next().unwrap(), iter.next().unwrap())
    };

    let mut a: std::collections::HashSet<usize> = std::collections::HashSet::new();
    for _ in 0..m {
        let mut line = String::new();
        std::io::stdin().read_line(&mut line).ok();
        a.insert(line.trim().parse().unwrap());
    }

    let mut dp = vec![0; n + 1];
    dp[0] = 1;
    for i in 0..(n + 1) {
        if i + 1 <= n && !a.contains(&(i + 1)) {
            dp[i + 1] = (dp[i + 1] + dp[i]) % 1000000007;
        }
        
        if i + 2 <= n && !a.contains(&(i + 2)) {
            dp[i + 2] = (dp[i + 2] + dp[i]) % 1000000007;
        }
    }

    println!("{}", dp[n]);
}