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() {
  int N, M;
  cin >> N >> M;
  vector<int> a(M);
  for (int i=0; i<M; i++){
    cin >> a.at(i);
  }
  
  vector<int> b(M+1);
  for (int i=0; i<M+1; i++){
    if (i==0){
      b.at(i) = a.at(i) - 1;
    }
    else if (i==M){
      b.at(i) = N - a.at(M-1) - 1;
    }
    else {
      b.at(i) = a.at(i) - a.at(i-1) - 1;
    }
  }
  
  int64_t S=1LL;
  for (int i=0; i<M+1; i++){
    if ((i!=0 && i!=M) && b.at(i)==0){
      cout << 0 << endl;
      break;
    }
    int64_t c, d, e;
    for (int j=0; j<b.at(i); j++){
      if (j==0){
        c=1;
      }
      else if (j==1){
        d=c;
        c=2;
      }
      else {
        e=d;
        d=c;
        c=d+e;
      }
    }
    if (b.at(i)==0){
      c=1;
    }
    S *= c;
    if (i==M){
      cout << S%1000000007 << endl;
    }
  }
}