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: Runtime Error)

import java.util.*;

public class Main{
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        
        int N = sc.nextInt();
        int M = sc.nextInt();
        int[] a = new int[M+2];
        for(int i = 0; i < M+2; i++){
            if(i == 0){
                a[i] = 0;
                continue;
            } 
            if(i == M+1){
                a[i] = N;
                continue;
            }
            a[i] = sc.nextInt();
        }
        long count = 1;
        int index = 1;
        while (index <= M+1){
            if(index > 1 && index != M+1) {
                if((a[index]-a[index-1]) == 1) {
                    count = 0;
                    break;
                }
            }
            if(index == 1 || index == M+1) {
                count *= stairs(a[index]-a[index-1]-1);
            }else {
                count *= stairs(a[index]-a[index-1]-2);
            }
            index++;
        }
        System.out.println((count % 1000000007));
        sc.close();
    }

    static long kaijou(int n) {
        if(n == 0)
            return 1;
        long kaijou = 1;
        for(int i = 1; i <= n; i++)
            kaijou *= i;
        return kaijou;
    }

    static long stairs(int N) {
        long stairs = 0;
        for(int i = 0; i <= (N/2); i++)
            stairs += ( kaijou(N-i) / (kaijou(i) * kaijou(N-2*i)));
        return stairs;
    }
}