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 ll long long
#define fi first
#define se second
#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 rep11(i, n) for (int i = 1; i < (int)(n); ++i)
#define repo(i, o, n) for (int i = o; i < (int)(n); ++i)
#define repm(i, n) for (int i = (int)(n)-1; i >= 0; --i)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define pb(n) push_back(n)
#define mp make_pair
#define MOD 1002000007
#define INF LONG_LONG_MAX

typedef pair<int, int> P;
int main(){
    long long n,m;
    cin>>n>>m;
    vector<long long> ans;
    vector<int>oks(n+1,true);
    rep(i,m){
        int a;
        cin>>a;
        oks[a]=false;
    }
    ans.push_back(1);
    if(oks[1]){
        ans.push_back(1);
    }
    else{
        ans.push_back(0);
    }
    long long p=0;
    for(int i=2; i<=n; i++){
        if(oks[i]){
            long long c1=ans[i-1];
            long long c2=ans[i-2];
            long long c=(c1+c2)%1000000007;
            ans.push_back(c);
            }
        else{
            ans.push_back(0);
        }
    }
    cout<<ans[ans.size()-1]<<endl;
}