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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int main() {
  int N,M;
  cin>>N>>M;
  set<int> a;
  rep(i,M) {
    int x;
    cin>>x;
    a.insert(x);
  }
  vector<uint64_t> dp(N+1);//i段目にいくまでに何通り移動法があるか。ans=dp[N]
  dp[0]=1;
  if (!(a.count(1))) dp[1]=1;
  int i=2;
  for(int i=2;i<=N;i++){
    if(!(a.count(i))){
      if(a.count(i-1)&&a.count(i-2)){//答えは0で確定
        cout<<0<<endl;
        return 0;
      }
      else if(a.count(i-1)){//i-2からくるしかない
        dp[i]=dp[i-2];
      }
      else if(a.count(i-2)){//i-1からくるしかない
        dp[i]=dp[i-1];
      }
      else dp[i]=dp[i-1]+dp[i-2];
    }
  }
  uint64_t y = pow(10,9)+7;
  cout<<dp[N]%y<<endl;  
  return 0;
}