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: Accepted)

using static System.Console;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Typical_Stairs2
{
    class Program
    {
        static void Main(string[] args)
        {
            //以下入力時の処理
            string input = ReadLine();            
            string[] array = input.Split(' ');
            var totalStep = int.Parse(array[0]);//零段目を抜いた段数の総数なので＋１
            var totalBrokenStep =int.Parse( array[1]);//壊れた段の総数
            var step = new long[totalStep + 1];//零段目も含めた段数分の箱を用意
            var brokenStep = new int[totalBrokenStep+2];//壊れた階段用の箱を用意
            for (int i = 0; i < totalBrokenStep ; i++) {
                brokenStep[i] = int.Parse(ReadLine()); //一行ずつ読み込み
            }//壊れた階段段数を記録
            step[0] = 1;//0段目は一通りなので
                        //以下」一段目以降の処理//
            int N = 1000000007;///定数
            int count = 0;//壊れた階段＝＝処理中の段数となったときに比較する壊れた段数を次に移すために使う
            for (int j = 1; j < totalStep+1; j++) {
                
                if ( j< 2)
                {
                    step[j] = step[j - 1] ;//totalStep==1の時step[j-2]は存在しない
                }
                else {
                    step[j] = step[j - 1] + step[j - 2]%N; //一つ前への行き方と二つ前への生き方の総和を代入           
                }
                if (j == brokenStep[count])
                { //j段目の値＝＝壊れた階段の時にstep[j]=0を代入
                    step[j] = 0;
                    count++;//処理する段数を次に移すうつす
                }
            }
            
            WriteLine(step[totalStep] % N);//結果を表示
            ReadKey();

        }
    }
}