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: Accepted)

#include<bits/stdc++.h>

using namespace std;
using lint = long long;
#define MOD 1000000007

int main(void)
{
    int N, M;
    //lint mod = 1000000007;
    cin >> N >> M;
    vector<int> a(M);
    for(int  i = 0; i < M; i++){
        cin >> a[i];
    }
    /*
    for(int i = 1; i < M; i++){
        if(a[i] - a[i-1] == 1){
            cout << 0 << endl;
            return 0;
        }
    }
    */
    
    vector<lint> dp(N+1);
    int cnt = 0;
    for(int i = 0; i <= N; i++){
        if(cnt < M && a[cnt] == i){
            dp[i] = 0;
            cnt++;
        }
        else if(i == 0 || i == 1){
            dp[i] = 1;
        }
        else{
            dp[i] = (dp[i-1] % MOD + dp[i-2] % MOD) % MOD;
        }
        //cout << cnt << " " << dp[i] << endl;
    }

    cout << dp[N] << endl;


    return 0;
}