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: Runtime Error)

import java.util.Scanner;



public class Main {

	public static void main(String[] args) {



		Solve s= new Solve();
		s.solve();

	}
}


class Solve{
	 Scanner sc = new Scanner(System.in) ;

	 void solve() {
		 boolean no=false;
		 int N=sc.nextInt();
		 int M=sc.nextInt();
		 int[] a=new int[M];
		 long res=0;
		 long sum=1;
		 for(int i=0;i<M;i++) {
			 a[i]=sc.nextInt();
			 if(i==0) {
				 sum*=fib(a[i]-1);
				 System.out.println(i+" 1 ");
				 if(M==1) {
					 sum*=N-a[i]-1;
				 }
			 }else if(i==M-1){
				 sum*=fib(N-a[i]-1);
				 System.out.println(i+" N ");
			 }else {
				 sum*=fib(a[i]-a[i-1]-2);
				 System.out.println(i+" other ");
			 }
			 if(sum==0) {
				 sum=1;
			 }
		 }

		 for(int i=0;i<M-1;i++) {
			 if(a[i]==a[i+1]-1) {
				 no=true;
				 i=M;
				 System.out.println(no);
			 }
		 }
		 if(no==false) {
			 res=sum;
		 }


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



	 public static int fib(int num) {
		 int[] dp = new int[10000];
		 dp[1]=1;
		 dp[2]=2;
		 for(int i=3;i<=num;i++) {
			 dp[i]=dp[i-1]+dp[i-2];
		 }
		 return dp[num];
	 }


	 }