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 <algorithm>
#include <vector>
#include <cmath>
using namespace std;

#define N 1000000007

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

    vector<long long int> dp(n+2, 0);
    dp[1] = 1;
    dp[2] = 2;
    for(int i = 0; i < m; i++) {
        dp[v[i]] = -1;
    }
    if(dp[1] == -1) {
        dp[2]--;
    }

    if(dp[1] == -1 && dp[2] == -1) {
        cout << 0 << endl;
        return 0;
    }

    for(int i = 3; i <= n+1; i++) {
        //cout << "i: " << i << " " << dp[i] << endl;
        if(dp[i] == -1) {
            //
            continue;
        }
        else {
            if(dp[i-1] != -1) {
                dp[i]+=dp[i-1];
            }
            if(dp[i-2] != -1) {
                dp[i]+=dp[i-2];
            }
        }
        dp[i]%=N;
    }

    /*
    for(int i = 0; i < n+1; i++) {
        cout << "i: " << i << " " << dp[i] << " " << endl;
    }
    cout << endl;
    */

    cout << dp[n] % N << endl;

    return 0;

}