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;

int main() {
  int64_t i=2,k=1,M,N;
  int64_t  count;
  cin>>N>>M;
  vector<int64_t> vec(M+1);
  vector<int64_t> v(2000000000);
  vector<int64_t> x(M+1);
  v.at(0)=1;
  v.at(1)=1;
  while(i<2000000000){
    v.at(i)=v.at(i-1)+v.at(i-2);
    i++;
  }
  for(int64_t j=0;j<M;j++){
    cin>>vec.at(j);
  }
  vec.at(M)=N+1;
  x.at(0)=v.at(vec.at(0)-1);
  while(k<M+1){
    if(vec.at(k)-vec.at(k-1)==1){
      x.at(k)=0;
    }
    else{
      x.at(k)=v.at(vec.at(k)-vec.at(k-1)-2);
    }
    k++;
  }
  count=x.at(0);
  for(int64_t m=1;m<M+1;m++){
    count*=x.at(m);
  }
  cout<<count%1000000007<<endl;
}