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: Runtime Error)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef map<int, int> mii;
typedef map<char,int> mci;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
typedef vector<pair<int,int>> vii;
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define in insert
#define sz(v) v.size()
#define mina(a,b) (a)=min((a),(b));
#define maxa(a,b) (a)=max((a),(b));
const int INF = 1e9 + 5;
void solve(){
    int n, m;  cin>>n>>m;
    vi v(n);
    for(int i = 0; i < m; i++) cin>>v[i];
    vi dp(n+1,0);
    int cur=0;
    if(v[cur]==1){
        cur++;
        dp[2]++;
    }
    else dp[1]=1;
    if(v[cur]==2){
        cur++;
    }
    else dp[2]=dp[1]+1;
    for(int i = 3; i<= n; i++){
        if(i==v[cur]){
            cur++;
            continue;
        }
        dp[i]=(dp[i-1]+dp[i-2])%1000000007;
    }

    //for(int it:dp ) cout<<it<<" ";
    cout<<dp[n];

}
int main(){
    int T=1;
    //cin>>T;
    for(int i = 0; i < T; i++) solve(); 
}