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;

const int64_t mod=1e9+7;

int main() {
  int N,M,i,x;
  cin >> N >> M;
  set<int> a;
  for(i=0; i<M; i++) {
    cin >> x;
    a.insert(x);
  }
  
  vector<int64_t> dp(N+1);
  dp.at(0)=1;
  
  for(i=1; i<=N; i++)
    if(!a.count(i)) {
      dp.at(i) += dp.at(i-1);
      if(i!=1)
        dp.at(i) += dp.at(i-2);
      if(dp.at(i) == 0) {
        cout << 0 << endl;
        return 0;
      }
      dp.at(i) %= mod;
    } else
      dp.at(i) = 0;
  
  cout << dp[N] << endl;
}