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 sc = new Scanner(System.in);
		int sum = sc.nextInt();
		int broken_sum = sc.nextInt();
		int[] array = new int[broken_sum + 1];

		for (int i=0; i<broken_sum; i++) {
			array[i] = sc.nextInt();
		}
		array[broken_sum] = sum + 1;

		long result = (fibonatch(array[0]))% 1000000007;

		for (int i=1; i<broken_sum+1; i++) {
			if (array[i] -  array[i-1] == 1) {
				result = 0;
				break;
			}
			result = (result * fibonatch(array[i] -  array[i-1] - 1) % 1000000007) % 1000000007;
		}


		System.out.println(result % 1000000007);
	}

	static int fibonatch(int num) {
		int tmp1 = 1;
		int tmp2 = 1;


		if (num < 3) {
			return 1;
		} else {
			int n = 0;
			for (int i = 3; i <= num; i ++) {
				n = tmp1 + tmp2;
				tmp1 = tmp2;
				tmp2 = n;
			}
			return n;
		}

	}

}