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: Runtime Error)

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

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(),
        )
    };

    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 step: Vec<i32> = if a[0] == 1 {
        vec![ 1, 0 ]
    } else {
        vec![ 1, 1 ]
    };
    
    for i in 2..n+1 {
        if a.contains(&i) {
            step.push( 0 );
        } else {
            let c = step[i-2] + step[i-1];
            if c > 1_000_000_007 {
                step.push( c % 1_000_000_007);
            } else {
                step.push( c );
            }
        }
    }
    
    println!("{}", step[n] );
}