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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<iostream>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<cmath>
#include<map>

using namespace std;

#define rep(i,N) for(int i=0;i<N;i++) //0から


typedef long long ll;




ll binary_search(vector<ll> a, ll n, ll key){
    ll right = n, left = -1;
    ll md = (right + left) / 2;

    while(right - left > 1){
        if(a[md] <= key){
            right = md;
        }else{
            left = md;
        }
        md = (right + left) / 2;
    }
    if(left == -1) return -1; //無い場合
    return right;
}

vector<ll> prime;
void Prime(ll n){ //線形篩,素数列挙
    vector<ll> p(n,0);
    p[0]=1;
    p[1]=1;
    for(ll i=2;i<n;i++){
        if(p[i]==0){
            prime.push_back(i);
            p[i]=i;
        }
        ll k=prime.size();
        for(ll j=0;j<k && i*prime[j]<n && prime[j]<=p[i];j++){
            p[i*prime[j]]=prime[j];
        }
    }
}

#define MOD ((ll)1e+9 + 7)




//cout<<fixed<<setprecision(10);
int main(){
    int n,m;
    cin>>n>>m;
    vector<int> a(m);
    rep(i,m){
        cin>>a[i];
        if(i!=0){
        }
    }


    vector<ll> cnt(n+1,0);
    cnt[0]=1;
    int c=0;
    rep(i,n+1){
        cnt[i]%=MOD;
        if(a[c]==i){
            c++;
            continue;
        }
        if(i+1<=n){
            cnt[i+1]+=cnt[i];
        }
        if(i+2<=n){
            cnt[i+2]+=cnt[i];
        }
    }

    cout<<cnt[n]<<endl;


    return 0;
}