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;
import java.util.Arrays;

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[n+1];//使えない階段の番号(インデックスの値に対応)
    int[] b = new int[n];//i+1段目までの場合の数
    boolean j0,j1,j2 ;
    int count = 0;

    for(int i=0; i<m; i++){
      a[sc.nextInt()] = -1;
    }
    b[0] = (a[1]==-1)?0:1;//1段目
    b[1] = (a[1]==-1)?1:2;//2段目
    for(int i=2;i<n;i++){
      if(a[i+1] == -1){continue;}
      b[i-1] = (a[i]==-1)? 0:b[i-1];
      b[i-2] = (a[i-1]==-1)?0:b[i-2];
      b[i] = b[i-1] + b[i-2];
      b[i] = b[i]%1000000007;
    }
    System.out.println(b[n-1]);


  }
}