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::collections::HashMap;

#[allow(unused_imports)]
use std::collections::HashSet;

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()
}

#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

const MOD: u64 = 1_000_000_007;

fn main() {
    let nm = read_vec::<usize>();
    let n = nm[0];
    let m = nm[1];

    let mut a : Vec<usize> = Vec::with_capacity(m + 1);
    for i in 0..m {
        a.push(read::<usize>());
    }
    a.push(n + 1);


    let answer = solve(n, m, &a);
    println!("{}", answer);
}

fn solve(n: usize,m: usize, a: &[usize]) -> u64 {

    let mut amap : HashMap<usize, u32> = HashMap::new();

    let mut from = 0;
    let mut max = 0;
    for k in a {
        if *k == from {
            return 0;
        }
        let d = k - 1 - from;
        if d > 0 {
            let count = amap.entry(d).or_insert(0);
            *count += 1;

            if d > max {
                max = d;
            }
        }
        from = k + 1;
    }

    let mut f0 = 0;
    let mut f1 = 1;
    let mut s : u64 = 1;
    for i in 1..max + 1 {
        let f = (f0 + f1) % MOD;
        
        if let Some(count) = amap.get(&i) {
            for p in 0..*count {
                s = (s * f) % MOD;
            }
        }

        f0 = f1;
        f1 = f % MOD;
    }


    return s;
}