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: Wrong Answer)

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;

#define INF 1000000007

int function(int x, int y){
    long long int memo[x];
    if(x == 0 || x==1){
        memo[0]=y; memo[1]=y;
        return y;
    }
    else if(x >= 2){
        memo[x]= (function(x-1, y)+function(x-2, y))%INF;
        return memo[x];
    }

}

int main(void){
    int N, M;
    cin >> N >> M;
    int check[M+2];
    check[0]=0; check[M+1]=N;
    long long int ans = 1;
    for(int i = 1; i < M+1; i++)
        cin >> check[i];

    for(int i = 1; i < M+2; i++){
        if(check[i]==check[i-1]+1 && i!=1 && i!=M+1){
            ans = 0;
            break;  
        }
        else if(i==1 || i==M+1)
            ans=function(check[i]-check[i-1]-1, ans);
        else if(i>1)
            ans=function(check[i]-check[i-1]-2, ans);
    }

   cout << ans << endl;
    return 0;
}