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<iostream>
#include<vector>
#define NMAX 100001

using namespace std;

int N, M;
vector<int> a;
int dp[NMAX];
int isOK[NMAX];

long long int fdp(long int i){
  if(dp[i]>0){
    return dp[i];
  }else{
    if(isOK[i]==1){
      dp[i]=0;
      return dp[i];
    }
    if(i==0||i==1){
      dp[i]=1;
      return dp[i];
    }
    dp[i]=(fdp(i-1)+fdp(i-2))%1000000007;
    return dp[i];
  }
}

int main(void){
  cin >> N >> M;
  for(int i=0;i<M;i++){
    long int tmp;
    cin >> tmp;
    isOK[tmp]=1;
  }
  cout << fdp(N);
  return 0;
}