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)

#include <bits/stdc++.h>
#include <math.h>
using namespace std;
int ope(int s, int t){
    return s+t;
}
int main(){
    int n,m;
    cin>>n>>m;
    vector<int> data;
    vector<int> num;
    for (int i=0; i<m; i++){
        int a;
        cin>>a;
        data.push_back(a);
    }
  
    for (int i=0; i<=n; i++){
        if (i==0){
            num.push_back(1);
        }
        if (i==1){
            if (data[0]==1){
                num.push_back(0);
            }
            else{
                num.push_back(1);
            }
        }
        if (i>=2){
            int k=ope(num[i-2],num[i-1]);
            
            for (int j=0; j<m; j++){
                if (data[j]==i){
                    k=0;
                }
            }
            num.push_back(k);
            
        }
    
    } 
    cout<<num[n]<<endl;
    
}