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.*;
    public class Main {// Main
        public static void main(String[] args){
            Scanner sc = new Scanner(System.in);
            int n = sc.nextInt();
            int m = sc.nextInt();
            long mod = 1000000007L;
            long ans = 1;
            int[] a = new int[m];
            int[] path = new int[n+1];
            
            int cool = 0;

            for(int i=0;i<m;i++) { a[i] = sc.nextInt(); path[a[i]]=1; 
               path[a[i]-1]=1;
                 path[a[i]-1]=1; 
            }   
            
            for(int i=0;i<m-1;i++) { if(a[i+1]-a[i]==1) { cool=1; break;}  }
           
            if(cool == 1 ) System.out.println(0);
            else{
                int count = 0;
                for(int i=0;i<=n;i++) { if(path[i]==0) count++; }
                if(path[1]==0) count--;
                if(path[n]==0) count--;
                int x = count/2;
                for(int i=1;i<=x;i++){
                    long val = 1;
                    for(int f=count-i+1;f<=count;f++){ val *= f; val %= mod; }
                    long me = 0;
                 //if(count-i+1>0)  me =  ((count-i+1)*(count-i+1))%mod; 
                    ans = (ans + val -me)%mod ;
                }
                System.out.println(ans);
            }
        }
    }