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: Accepted)

import java.util.*;
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+1];
   for(int i=0;i<M;i++) a[i] = sc.nextInt();
   long[] dp = new long[N+1];
   int count =0;
   if(N==1) System.out.println(1);
   else
   {

   dp[1]=1;
   if(a[count]==1)
   {
     dp[1]=0;
     count++;
   }
   dp[2]=1+dp[1];
   if(a[count]==2)
   {
     dp[2]=0;
     count++;
   }
   for(int i=3;i<=N;i++)
   {
     if(a[count]==i) 
     {
       dp[i]=0;
       count++;
     }
     else dp[i]=(dp[i-1]+dp[i-2])%1000000007;
   }
   System.out.println(dp[N]);
   }
 }
}