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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll nmax = 1000000007;

int main() {
  int n,m;
  cin >> n >> m;
  vector<ll> dp(n+1,0);
  vector<int> a(m);
  for(int i=0;i<m;i++){
    cin >> a.at(i);
  }
  
  auto itr = find(a.begin(),a.end(),1);
  if(itr == a.end()){
    dp.at(1) = 1;
  }
  
  itr = find(a.begin(),a.end(),2);
  if(itr != a.end()){
    dp.at(2) = 0;
  }else{
    dp.at(2) = dp.at(1) + 1;
  }
  
  for(int i=3;i<n+1;i++){
    itr = find(a.begin(),a.end(),i);
    if(itr == a.end()){
      dp.at(i) = (dp.at(i-1) + dp.at(i-2)) % nmax;
    }else{
      dp.at(i) = 0;
    }
  }
	
  cout << dp.at(n) << endl;
  return(0);
}