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: Wrong Answer)

use std::io;
use std::io::BufRead;

const MOD: u64 = 1_000_000_007;


fn main() {
    let stdin = io::stdin();
    let mut stdin = stdin.lock();
    
    let (n,m) = {
        let mut s = String::new();
        stdin.read_line(&mut s).unwrap();
        let mut s = s.trim().split_whitespace();
        (
            s.next().unwrap().parse::<usize>().unwrap(),
            s.next().unwrap().parse::<usize>().unwrap(),
        )
    };
    if m == 0 { println!("{}", fibonacci(n).last().unwrap()); return; }

    let a: Vec<usize> = {
        let mut a = Vec::new();
        for _ in 0..m {
            let mut s = String::new();
            stdin.read_line(&mut s).unwrap();
            a.push(
                s.trim().parse::<usize>().unwrap()
            );
        }
        a
    };


    let mut lag: usize = a[0]+1;
    
    for i in 0..m-1 {
        let l = a[i+1] - a[i];
        
        if l == 1 {
            println!("0");
            return;
        }

        if l > lag { lag = l; }
    }
    if n-a[m-1] > lag {
        lag = n-a[m-1];
    }

    let phi: Vec<u64> = fibonacci( lag-2 );

    let mut prod: u64 = phi[a[0]-1];
    for i in 0..m-1 {
        println!("phi = {}", phi[ a[i+1] - a[i] - 2 ] );
        
        prod *= phi[ a[i+1] - a[i] - 2 ];
        // if i%3 == 0 {
        //     prod = prod % MOD;
        // }
    }

    println!("{}", ( prod * phi[n - a[m-1] - 1])%MOD);
}



fn fibonacci( n: usize ) -> Vec<u64> {
    let mut phi: Vec<u64> = vec![ 1, 1 ];

    for i in 2..n+1 {
        let sum = phi[i-2] + phi[i-1];
        phi.push(sum);

        // if i%16 == 0 {
        //     phi[i] %= MOD;
        //     phi[i-1] %= MOD;
        // }
    }
    phi
}