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;
typedef long long ll;
#define pb pushback
#define fr(i,n) for(int i=0;i<n;i++)
#define ifr(i,n) for(int i=n-1;i>=0;i--)


int main() {
    int n,m;
	cin >> n >>m;
    long a[m+1];
    long ans[n+1];
    int j=0;
    int de =1000000007;
   fr(i,m){
     cin >> a[i];
   }
   a[m]=0;
   ans[0]=1;
   if(a[0]==1){
     ans[1]=0;
     j =1;
   }else{
     ans[1]=1;
   }
    for(int i=2;i<n+1;i++){
     if(a[j]==i){
       ans[i]=0;
       j++;
     }else{
       ans[i]=ans[i-2]+ans[i-1];
       ans[i]%=de;
     }
   }
    cout << ans[n] << endl;
}