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 std = new Scanner(System.in);
		 long n = std.nextInt();
		 int m = std.nextInt();
		 long[] a = new long[m];
		 boolean ok = true;
		 for(int i=0;i<m;i++) {
			 a[i] = std.nextInt();
			 if(i>=1) {
				 if(a[i]-a[i-1]==1) {
					 ok = false;
				 }
			 }
		 }
		 if(ok) {
			 if(m >=1) {

				 long sum =a[0]-1;
				 if(sum==0) {
					 sum=1;
				 }
				 for(int i=1;i<m;i++) {
					 long dis = a[i]-a[i-1]-2;
					 sum = sum*fib(dis+1);
					 sum = sum%1000000007;
				 }
				 System.out.println((sum*fib(n-a[m-1]-1+1))%1000000007);
			 }
			 else {
				 System.out.println(fib(n+1));
			 }
		 }



		 else {
			 System.out.println(0);
		 }
	 }

	    private static long fib(long num) {//n段の確率(-1すること

	        if (num == 0) {
	            return 0;
	        }
	        if (num == 1) {
	            return 1;
	        }
	        return fib(num - 1) + fib(num - 2);
	    }
}