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: Time Limit Exceeded)


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

public class Main {

	public static int N;
	public static int[] stair;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		N = sc.nextInt();
		int M = sc.nextInt();
		List<Integer> broken = new ArrayList<>();
		for(int i = 1; i < M + 1; i++) {
			broken.add(sc.nextInt());
		}
		stair = new int[N + 1];
		for(int i = 1; i < N + 1; i++) {
			stair[i] = 0;
		}
		for(Integer e : broken) {
			stair[e] = 1;
		}
		long ans = Main.Stair(0);
		System.out.println(ans % 1000000007);
	}

	private static long Stair(int n) {
		long count = 0;
		if(n >= N - 1) {
			count = 1;
		} else if(n < N - 1) {
			if(stair[n + 1] == 0) {
				count += Main.Stair(n + 1);
			}
			if(stair[n + 2] == 0) {
				count += Main.Stair(n + 2);
			}
		}
		return count;
	}

}