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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System; 
using System.Text; 
using System.Collections.Generic;
using System.Linq; 
using E = System.Linq.Enumerable; 

class Program { 
     static int[] stairs;
     static void Main(string[] args) {
     //入力を受け取る 
     int[] NM = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
     int N =　NM[0];
     int M =　NM[1];
     stairs = new int[N+1];
     
     stairs[0] = 1;
     stairs[1] = 1;
     
     for(int i = 2; i < N+1 ;i++){
          stairs[i] = -1;
     }
      
     for(int i = 1; i < M+1 ;i++){
          var num = int.Parse(Console.ReadLine());
          stairs[num] = 0;
     } 
      Console.WriteLine(CountWay(N)); }
      
     static int CountWay(int dan){
         if(stairs[dan] != -1){
             return stairs[dan];
         }else{
             stairs[dan] = CountWay(dan-1)+CountWay(dan-2);
             return stairs[dan];
         }
     }
     }