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 (1.42.0) (judge verdict: Accepted)

use proconio::{input, fastout};
use std::usize::MAX;

const MOD: usize = 1_000_000_007;

fn dp(step_cnt: &mut Vec<usize>, steps: usize) -> usize {
    if step_cnt[steps] != MAX { return step_cnt[steps]; }
    step_cnt[steps] = ( dp(step_cnt, steps-1) + dp(step_cnt, steps-2) ) % MOD;
    step_cnt[steps]
}

#[fastout]
fn main() {
    input! {
        n: usize, m: usize,
        broken: [usize; m],
    }

    let mut step_cnt: Vec<usize> = vec![MAX; n+4];
    step_cnt[0] = 1;
    step_cnt[1] = 1;
    step_cnt[2] = 2;

    let mut ans;
    let mut last_idx = 0;
    let mut it = broken.iter();
    let f = it.next();
    if f == None {
        ans = dp(&mut step_cnt, n-last_idx);
    } else {
        let i = f.unwrap();
        ans = dp(&mut step_cnt, i-last_idx-1);
        last_idx = *i;
        for i in it {
            if i-last_idx==1 { ans=0; break }
            ans = ans * dp(&mut step_cnt, i-last_idx-2) % MOD;
            last_idx = *i;
        }
        ans = ans * dp(&mut step_cnt, n-last_idx-1) % MOD;
    }

    println!("{}", ans);
}