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>
#define rep(i, n, k) for(int i=k; i<n; ++i)
using namespace std;

int main(){
  long n, m;
  long long int ans = 1;
  cin >> n >> m;
  vector<long> a(m+2);
  vector<long> num(n+1);
  a.at(0) = -1;
  a.at(m+1) = n+1;
  rep(i, m+1, 1) cin >> a.at(i);
  
  num.at(0) = 0;
  num.at(1) = 0;
  num.at(2) = 1;
  rep(i, n+1, 3) num.at(i) = num.at(i-1) + num.at(i-2);
  
  rep(i, m+2, 1){
    int x = a.at(i) - a.at(i-1);
    ans *= num.at(x);
    ans = ans % 1000000007;
  }
  cout << ans << endl;
}