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)



#![allow(non_snake_case)]

use std::cmp::*;

fn read_line() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn read_vec() -> Vec<i32> {
    read_line()
        .split_whitespace()
        .map(|e| e.parse().unwrap())
        .collect()
}

fn read_int_tuple() -> (i32, i32) {
    let v = read_vec();
    (v[0], v[1])
}

fn main() {
    let (N, M) = read_int_tuple();
    let N = N as usize;
    let m = 1e9 as i32 +7;
    let mut dp = vec![0; N+1];
    for i in 0..M {
        let a: usize = read_line().parse().unwrap();
        dp[a] = -1;
    }
    dp[0] = 1;
    dp[1] = if dp[1] == -1 {-1} else {1};
    for i in 2..N+1 {
        if dp[i] != -1 {
            dp[i] = max(0, dp[i-1]) + max(0, dp[i-2]);
            dp[i] %= m;
        }
    }
    println!("{}", dp[N]);
}