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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n, m;
    cin >> n >> m;
    vector<int> a(m);
    if (m != 0){
        for (int i = 0;i < m;i++){
        cin >> a[i];
        a[i]--;
        }
    }

    int c = 0;
    vector<bool> s(n,true);

    if (m != 0){
        for (int i = 0; i < n; i++){
            if (i == a[c]){
                s[i] = false;
                c++;
            }
        }
    }
    
    vector<long long> ans(n,0);

    for (int i = 0; i < n; i++){
        if (i == 0){
            if (s[i] == false){
                ans[i] = 0;
            }
            else{
                ans[i] = 1;
            }
        }
        else if (i == 1){
            if (s[1] == false){
                ans[i] = 0;
            }
            else{
                ans[i] = ans[i-1] + 1;
            }
        }
        else{
            if (s[i] == false){
                ans[i] = 0;
            }
            else{
                ans[i] = ans[i-1] + ans[i-2];
                ans[i] %= 1000000007;
            }
        }
    }

    cout << ans[n-1] << endl;
}