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<iostream>
#include<math.h>
#include<vector>
#include<algorithm>
#include<numeric>
using namespace std;

vector<int> memo(100010,-1);
int p=pow(10,9)+7;
int func(int i){
    if(i==-1)return 0;
    if(i<=1)return 1;
    if(memo[i]!=-1)return memo[i];

    return memo[i]=(func(i-1)%p+func(i-2)%p)%p;
}

int main(){
    int N,M,a[100010];
    long long da[100010];
    cin>>N>>M>>a[0];

    if(M!=0){
    da[0]=a[0]-1;
    
    for(int i=1;i<M;++i){
        cin>>a[i];
        da[i]=a[i]-a[i-1]-2;
    }
    da[M]=N-a[M-1]-1;

    long long x=1;
    for(int m=0;m<=M;++m){
         x=(x*func(da[m]))%p;
    }

    cout<<x<<endl;}
    else cout<<func(N)<<endl;
    
    return 0;

}