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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define ALL(a)  (a).begin(),(a).end()
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;

#define MOD 1000000007

vector<ll> fib(100001, -1);

ll res(ll N, set<ll> a){
  if(fib.at(N) != -1) return fib.at(N);

  if(N == 1) return fib.at(N) = 1;
  if(N == 0) return fib.at(N) = 1;

  return fib.at(N) = (res(N-1, a) + res(N-2, a)) % MOD;
}

int main(){
  ll N, M;
  cin >> N >> M;
  set<ll> a;
  for(int i = 0; i < M; i++){
    ll aa;
    cin >> aa;
    a.insert(aa);
    fib.at(aa) = 0;
  }
  cout << res(N, a) << endl;
}