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: Time Limit Exceeded)

import java.util.Scanner;
public class Main {
	static int m;
	static int n;
	static int[] a;
	static long [] map;
	    public static void main(String[]args){
	        Scanner scan = new Scanner (System.in);
	        n = scan.nextInt();
	        m = scan.nextInt();
	        map = new long [n+1];
	        for(int i = 0;i < n+1;i++) {
	        	map[i] = -1;
	        }
	        if(m == 0) {
	        	a = new int[1];
	        	a[0] = -1;
	        }else {
	        	a = new int[m];
	        	for(int i = 0; i < m;i++) {
	        		a[i] = scan.nextInt();
	        	}
	        }
	        Long ans = downSteps(n);
	        System.out.println(ans%1000000007);
	}
	    public static long downSteps (int Step) {
	    	if(map[Step] > -1) {
	    		return map[Step];
	    	}
	    	boolean nextStepExistence = true;
	    	boolean twoDownStepExistence = true;
	    	int nextStep = Step-1;
	    	int twoDownStep = Step-2;
	    	long pattern = 1;
	    	for(int i = 0; i < a.length;i++) {
	    		if(nextStep < 0||a[i] == nextStep || !nextStepExistence) {
	    			nextStepExistence = false;
	    		}
	    		if(twoDownStep < 0 || a[i] == twoDownStep || !twoDownStepExistence) {
	    			twoDownStepExistence = false;
	    		}
	    	}
	    	if(nextStepExistence) {
	    		pattern = downSteps(nextStep);
	    	}
	    	if(twoDownStepExistence) {
	    		if(nextStepExistence) {
	    			pattern += downSteps(twoDownStep);
	    		}else {
	    			pattern = downSteps(twoDownStep);
	    		}
	    	}
	    	if(!twoDownStepExistence&&!nextStepExistence&&Step != 0) {
	    		pattern = 0;
	    	}
	    	pattern = pattern % 1000000007;
	    	map[Step] = pattern;
	    	return pattern;
	    }
}