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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;
import java.util.ArrayList;

public class Main{

    static Long memo[];
    public static void main(String args[]){

        Scanner scan = new Scanner(System.in);

        int N = Integer.parseInt(scan.next());
        int M = Integer.parseInt(scan.next());
        ArrayList<Integer> E = new ArrayList<Integer>();

        for(int i=0;i<M;i++)
            E.add(Integer.parseInt(scan.next()));

        memo = new Long[N+1];
        for(int i=0;i<N+1;i++) memo[i] = 0L;

        System.out.println(getConbination(N, E) % 1000000007);
    }

    static Long getConbination(Integer N,ArrayList E){

        if(memo[N-1] != 0)
            return memo[N-1];

        if(N == 1){
            if(E.indexOf(1L) == -1){
                memo[0] = 1L;
                return 1L;
            }else{
                memo[0] = 0L;
                return 0L;
            }
        }else if(N == 2){
            if(E.indexOf(1L) == -1){
                memo[1] = 2L;
                return 2L;
            }else{
                memo[1] = 1L;
                return 1L;
            }
        }

        if(E.indexOf(N-1) != -1 && E.indexOf(N-2) != -1){
            return 0L;
        }else if(E.indexOf(N-1) != -1){
            return getConbination(N-2, E);
        }else if(E.indexOf(N-2) != -1){
            return getConbination(N-1, E);
        }else{
            Long memo_ = getConbination(N-1, E) + getConbination(N-2, E);
            memo[N-1] = memo_;
            return memo_;
        }
    }
}