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 rep1(i,N) for(int i=1;i<int(N);++i)
#define all(a) (a).begin(),(a).end()		//sort(all(vi S)) sort(all(string S))
#define print(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<_<<", "; cerr<<"]"<<endl; }
#define printpair(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<"{"<<_.first<<","<<_.second<<"}"<<", "; cerr<<"]"<<endl; }

using P = pair<int, int>;		//P.first, P.second
typedef long long ll;
typedef vector<int> vi;
typedef set<int> seti;
typedef vector<string> vs;

const int MOD = 1e9+7;
const int INF = 1e9;

ll dp[100005];
int breaked[100005];
int main() {
	ll N,M;
	cin>>N>>M;
	int a;
	rep(i,M){
		cin>>a;
		breaked[a] = 1;
	}
	dp[0] = 1;
	rep(i,N+1){
		if(!(breaked[i+1])){
			if(i-1>=0){
				dp[i+1]+= dp[i]+dp[i-1];
				dp[i+1]%=MOD;
			}
			else{
				dp[i+1]+= dp[i];
				dp[i+1]%=MOD;
			}
		}
/*		if(!(breaked[i+2])){
			if(!(breaked[i+1])){
				dp[i+2]+= dp[i+1];
				dp[i+2]%=MOD;
			}
			if(!(breaked[i])){
				dp[i+2]+= dp[i];
				dp[i+2]%= MOD;
			}
		}*/
	}
	rep(i,N+1){
		cerr<<dp[i]<<" ";
	}
	cerr<<endl;
	ll ans = dp[N]%MOD;
	cout<<ans<<endl;
}