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 <fstream>
#include <vector>
#include <cstring>


#define REP(i, n) for (long long i = 0; i < (n); i++) 
#define ll long long

using namespace std;

int main()
{
    //    std::ifstream in("input.txt");
//    std::cin.rdbuf(in.rdbuf());
    ll N,M,res;
    cin >> N >> M;

    ll dp[N+1];
    std::memset(dp, 0, sizeof dp);

    bool dan[N+1];
    std::memset(dan, true, sizeof dan);

    ll tmp;
    REP(i,M)
    {
        cin >> tmp;
        dan[tmp] = false;
    }

    dp[0] = 1;//1通り
    if(dan[1] == true)dp[1] = 1;//1段目が壊れてなければ1通り

    for(ll i = 2; i <= N; i++)
    {
        if(dan[i] == true)
        {
        dp[i] += dp[i-1]; 
        dp[i] += dp[i-2]; 
        }
        //else dp[i] = 0;
    }
    //1000000007
  
    dp[N] = dp[N] % 1000000007;
    cout << dp[N] << endl;

    return 0;
}