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 N = sc.nextInt();
		int M = sc.nextInt();
		int[] A = new int[M];
		for(int i = 0 ; i < M ; i++){
			A[i] = sc.nextInt();
		}
		sc.close();

		long INF = 1000000007;
		int[] B = new int[M+1];
		long[] dp = new long[N+1];
		boolean flag = true;

		dp[0] = 1;
		dp[1] = 1;

		if(M == 0){
			for(int j =2 ; j <= N ; j++){
				dp[j] = (dp[j-1] + dp[j-2])%INF;
			}
			System.out.println(dp[N]);
		}
		else{
			B[0] = A[0]-1;
			if(A[0]-1 == 0){
				B[0] = 1;
			}
			B[M] = N-(A[M-1]+1);
			if(B[M] == 0){
				B[M] = 1;
			}
			for(int i = 0 ; i < M-1 ; i++){
				B[i+1] = (A[i+1]-1) - (A[i]+1);
				if(B[i+1]<0){
					flag = false;
					break;
				}
			}
			long ans = 1;
			if(flag){
				for(int i = 0 ; i <=M ; i++){
					//System.out.println(B[i]);
					if(dp[B[i]] != 0){
						ans *= dp[B[i]]%INF;
					}
					else{
						for(int j =2 ; j <= B[i] ; j++){
							dp[j] = (dp[j-1] + dp[j-2])%INF;
						}
						ans *= dp[B[i]];
					}
				}
				System.out.println(ans%INF);
			}
			else{
				System.out.println("0");
			}
		}
	}
}