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 <iostream>
#include <vector>
using namespace std;
typedef long long ll;
const ll mod =1000000007;
int main(){
  ll n,m;
  cin >> n >> m;
  vector<ll> broken(n);
  vector<ll> dp(n+2,0);
  dp.at(n+1)=0;
  dp[n]=1;
  
  for (int i=0;i<m; i++) {
    int a;
    cin >> a;
    broken.at(a)=1;
  }
  for (int j=n-1; j > -1; --j){
    if (broken[j]==1) continue;
    dp[j]=dp[j+1]%mod+dp[j+2]%mod;
   }
  //for (int h=0;h < n;h++) cout <<broken[h] << " ";
  //printf("\n");
  //for (int i=0;i<n+2;i++) cout << dp[i] << " ";
  //printf("\n");
  cout << dp[0] << endl;
  return 0;
}