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>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep2(i,s,n) for(int i=(int)(s);i<(int)(n);i++)
typedef long long ll;
const ll MOD = 1e+9+7;

int main(){
    int N,M;
    cin >> N >> M;
    vector<int> a(M);
    ll ans=-1;
    rep(i,M){
        cin >> a[i];
        if(i>0&&a[i]==a[i-1]+1){
            ans=0;
        }
    }

    //フィボナッチ
    vector<ll> f(N+1);
    f[0]=1;
    f[1]=1;
    rep(i,N-1){
        f[i+2]=f[i+1]+f[i];
        f[i+2]%=MOD;
    }


    if(ans==0){
        cout << ans <<endl;
    }else if(M>0){
        ans=1;
        ans*=f[a[0]-1];
        rep(i,M-1){
            ans*=f[a[i+1]-a[i]-2];
            ans%=MOD;
            
        }
        ans*=f[N-a[M-1]-1];
        ans%=MOD;
        cout << ans << endl;
    }else{
        ans = f[N];
        cout << ans << endl;
    }
}