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.Scanner;
public class Main {

	static long ln = 1000000007;
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		long fibdp[] = new long[n];
		fibdp[0] = 1;
		fibdp[1] = 1;
		for(int i=2;i<n;i++) {
			fibdp[i] = fibdp[i-1] + fibdp[i-2];
		}
		int a[] = new int[m];
		for(int i=0;i<m;i++) {
			a[i] = sc.nextInt();
		}
		long ans = fib(a[0] -1, fibdp)% ln;
		for(int i=1;i<m;i++) {
			ans = ans * fib(a[i]-a[i-1]-2, fibdp)% ln;
		}
		ans = ans * fib(n - a[m-1] -1, fibdp)% ln;
		
		System.out.print(ans);
	}
	
	public static long fib(int i, long[] fibdp) {
		if(i<0)
			return 0;
		else 
			return fibdp[i];
	}
}