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 <iostream>
#include <vector>

using namespace std;

int main()
{
  size_t n, m;
  cin >> n >> m;
  
  vector<size_t> holes(m);
  for(size_t i = m; i != 0; --i)
    cin >> holes[i-1];

  n += 1;
  vector<unsigned long long> steps(n);
  for(size_t i = 0; i < n; ++i)
  {
    if(m > 0 && holes[holes.size()-1] == i)
    {
      holes.pop_back();        
      steps[i] = 0;
      continue;
    }
    
    if(i < 2)
    {
    	steps[i] = 1;
      	continue;
    }
    
    steps[i] = (steps[i-1] + steps[i-2]) % (unsigned long long)1000000007;
  }
  cout << steps[steps.size()-1] << endl;
  
  return 0;
}