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.*;
import static java.lang.System.*;

public class Main{
  public static void main(String[] args){
	  Scanner sc = new Scanner(System.in);
	  
	  int N = sc.nextInt();
	  int M = sc.nextInt();
	  long [] DP = new long[N + 1]; //動的計画法

	  for(int i=0; i<M; i++) {
		  int n = sc.nextInt();
		  DP[n + 1] = -1;
	  }
	  DP[0] = 1;
	  if(DP[1] != -1)DP[1] = 1;
	  else DP[1] = 0;
	  
	  for(int i=2; i<=N; i++) {
		  if ( DP[i] == -1) DP[i] = 0;
		  else {
			  DP[i] = DP[i-1] + DP[i-2];
		  }
	  }
	  out.println(DP[N] % 1000000007);
  }
}