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)

use std::vec::Vec;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
  let MOD: u64 = 1_000_000_007;
  let (n, m) = {
    let v: Vec<usize> = read_vec();
    (v[0], v[1])
  };

  let mut a: Vec<u64> = Vec::new();
  for _ in 0..m {
    let at = read();
    a.push(at);
  }
  let mut dp: Vec<u64> = vec![0; n+1];
  dp[0] = 1;
  dp[1] = 1;
  for i in 2..n+1 {
    let bn1 = match a.binary_search(&((i-1) as u64)) {
      Ok(i) => true,
      Err(err) => false,
    };

    let bn2 = match a.binary_search(&((i-2) as u64)) {
      Ok(i) => true,
      Err(err) => false,
    };
    if !bn1 && !bn2 {
      dp[i] = dp[i-1] + dp[i-2] % MOD;
    }
    else if bn1 && !bn2 {
      dp[i] = dp[i-2] % MOD;
    }
    else if !bn1 && bn2 {
      dp[i] = dp[i-1] % MOD;
    }

  }

  println!("{:?}", dp[n] % MOD);
}