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;

public class Main{
    public static void main(String[] args){
        Scanner stdIn = new Scanner(System.in);

        int n = stdIn.nextInt();
        int m = stdIn.nextInt();
        int[] a = new int [m];
        for(int i=0; i<m; i++){
            a[i] = stdIn.nextInt();
        }

        long[] fibo = new long[n+1];
        fibo[0] = 1;
        if(fibo.length > 1){
            fibo[1] = 1;
        }
        if(fibo.length > 2){
            fibo[2] = 2;
        }
        for(int i=3; i<fibo.length; i++){
            fibo[i] = fibo[i-2] + fibo[i-1];
        }

        if(a.length == 0){
            System.out.println(fibo[n]%=1000000007l);
            return;
        }

        for(int i=0; i<a.length-1; i++){
            if(a[i+1]-a[i] == 1){
                System.out.println(0);
                return;
            }
        }

        long sum = 1;
        for(int i=0; i<a.length; i++){
            if(i == 0){
                if(a[i] == 1){
                    sum *= 1;
                }else{
                    sum *= fibo[a[i]-1];
                }
            }else{
                sum *= fibo[a[i]-a[i-1]-2];
            }
            if(sum >= 1000000007l){
                sum %= 1000000007l;
            }
        }
        sum *= fibo[n-a[m-1]-1];

        System.out.println(sum%=1000000007l);
    }
}