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);

		long x=1000000007;

		int N=sc.nextInt();

		long[] dp=new long[N+1];
		for(int i=0;i<N+1;i++) dp[i]=0L;
		dp[0]=1L;

		int M=sc.nextInt();
		for(int i=0;i<M;i++) dp[sc.nextInt()]=-1L;

		for(int i=0;i<N;i++) {
			if(dp[i]==-1) continue;
			if(dp[i+1]!=-1) dp[i+1]=(dp[i+1]+dp[i])%x;
			if(i+2<=N  && dp[i+2]!=-1) dp[i+2]=(dp[i+2]+dp[i])%x;
			
		}

		System.out.println(dp[N]%x);

		sc.close();
	}

}