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: Accepted)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		long[] a = new long[M];
		
		long c = 1000000007;
		
		long[] fib = new long[N + 2];
		fib[0] = 1;
		fib[1] = 1;
		fib[2] = 2;
		
		for(int i = 3; i <= N + 1; i++) {
			fib[i] = fib[i - 1] + fib[i - 2];
			fib[i] = fib[i] % c;
		}
		 
		if(M == 0) {
			System.out.println(fib[N]);
			return;
		}
		
		for(int i = 0; i < M; i++) {
			a[i] = sc.nextInt();
			if(i > 0 && a[i] - 1 == a[i - 1]) {
				System.out.println(0);
				return;
			}
		}
		
		int now = 0;
		int x = 0;
		long sum = 1;
		int temp = 0;
		
		while(temp < N) {
			now++;
			temp++;
			if(x < M) {
				if(temp == a[x]) {
					sum = sum % c * fib[now - 1] % c;
					x++;
					now = -1;
				}
			}
		}
		sum = sum % c * fib[now] % c;
		System.out.println(sum);
	}
}