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 Java (OpenJDK 11.0.6) (judge verdict: Runtime Error)

import java.util.*;

public class Main{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int m = scan.nextInt();
        boolean[] a = new boolean[n];
        for (int i=0;i<m;i++){
            int t = scan.nextInt();
            a[t-1] = true;
        }
        long sum = 0;
        long[] g = new long[n];
        g[0] = 1;
        g[1] = 1;
        if (a[0] == true){
            g[1] = 0;
        }
        sum = find(2, n, a, g);
        long q = 1000000007;
        System.out.println((long) sum % q);
    }

    static long find(int current, int n, boolean[] a, long[] g){
        
        if (current == n){
            return g[n-1]+g[n-2];
        }
        
        if (a[current-1] == true){
            g[current] = 0;
        }
        else{
            g[current] = g[current-1] + g[current-2];
        }     
        long sum = 0;
        sum = find(current + 1, n, a, g);
        return sum;
    }
}