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.math.BigInteger;

public class Main {
	static int[] breaks;
	static int num;
	static BigInteger[] patterns;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		num = sc.nextInt();
		int m = sc.nextInt();
		breaks = new int[m];
		patterns = new BigInteger[num + 3];
		for(int i = 0; i < m; i++){
			breaks[i] = sc.nextInt();
		}
		System.out.println((pattern(0).longValue() % 1_000_000_007));
	}
	static BigInteger pattern(int p){
		if(patterns[p] != null){
			return patterns[p];
		}
		if(p == num){
			patterns[p] = BigInteger.ONE;
			return patterns[p];
		}
		if(p > num){
			patterns[p] = BigInteger.ZERO;
			return patterns[p];
		}
		for(int i = 0; i < breaks.length; i++){
			if(p == breaks[i]){
				patterns[p] = BigInteger.ZERO;
				return patterns[p];
			}
		}
		patterns[p] = pattern(p + 1).add(pattern(p + 2));
		return patterns[p];
	}
}