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>
using namespace std;
#define rep(i,n) for(int i = 0;i<n;i++)
using P = pair<int,int>;
using ll = long long;

int main() {
  int n,m;
  cin>>n>>m;
  vector<ll> v(m);
  rep(i,m){
    cin>>v.at(i);
  }
  ll INF = 1000000007;
  vector<ll> h(n+1,0);
  h.at(0) = 1;
  if (v.at(0) != 1){
    h.at(1) = 1;
  }
  for(ll i = 2;i<n+1;i++){
    h.at(i) = h.at(i-1) + h.at(i-2);
    h.at(i) = h.at(i) % INF;
    rep(j,m){
      if (i == v.at(j)){
        h.at(i) = 0;
      }
    }
  }
  cout << h.at(n) % INF << endl;
}