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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize ("-ffloat-store")  
#pragma GCC optimize ("-fno-defer-pop")
#define all(a) a.begin(),a.end()
#define ll long long int
#define ld long double
ll power(ll a,ll b,ll m){ if(b==0) return 1; if(b==1) return a%m; ll t=power(a,b/2,m)%m; t=(t*t)%m; if(b&1) t=((t%m)*(a%m))%m; return t;}
ll modInverse(ll a, ll m) { return power(a, m-2, m); }
#define ps push_back
#define fs first
#define sc second
#define N 3000005
#define endl "\n"
#define mod 1000000007
ll n,m;
map<ll,ll> mp;
ll dp[100005];
ll maxval(ll i){
	if(i>n) return 0;
	if(i==n){
		return 1;
	}
	if(mp.find(i)!=mp.end()){
		return 0;
	}
	if(dp[i]!=-1){
		return dp[i];
	}

	ll a,b;
	a=maxval(i+1)%mod;
	b=maxval(i+2)%mod;
	return dp[i]=(a+b)%mod;
}
int main(){
	
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	
	ll i,j,k,l;
	cin>>n>>m;
	memset(dp,-1,sizeof(dp));
	for(i=1;i<=m;i++){
		cin>>k;
		mp[k]=1;
	}
	if(mp.find(n)!=mp.end()){
		cout<<"0\n";
		return 0;
	}

	k=maxval(0)%mod;
	cout<<k;
	return 0;
}