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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>//復習必須
#include<iostream>
#include<vector>
#include <cmath>
#include <map>
#include <algorithm>
#include <string>
#define rep(i, n) for (int i = 0; i < n; ++i)
using ll = long long;
using namespace std;
#define P pair<int, int>
int main () {
    ll N , M ;
    cin >> N >> M ;
    
    vector<bool>oks(N + 1 , true) ;
    for(int i=0;i<M;++i){
    int a;cin>>a;
    oks[a]=false;
    }

   vector<ll>dp(N+1);
   dp[0]=1;
   for(ll now=0;now<N;++now){
   for(ll next=now+1;next<=min(N,now+2);next++){
   if(oks[next]){
   dp[next]+=dp[now];
   dp[next]%=1000000007;
   }
   }
   }
  
   cout<<dp[N]<<endl;

    
    


   
    return 0 ;
}