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

typedef long long ll;

#define int ll
#define all(v) v.begin(), v.end()
#define reps(__i, a, b) for (int __i = a; i < b; i++)
#define rep(__i, n) reps(__i, 0, n)

const ll INF = (1ll << 60);
const ll MOD = (ll)1e9 + 7;

signed main()
{
    int n, m;
    int a[100000];
    cin >> n >> m;
    for (int i = 0; i < m; i++)
        cin >> a[i];

    set<int> s;
    for (int i = 0; i < m; i++)
        s.insert(a[i]);

    int dp[100001] = {1, 1};
    for (int i = 2; i <= n; i++)
        dp[i] = (dp[i - 1] + dp[i - 2]) % MOD;

    // for (int i = 0; i <= n; i++)
    //     cout << dp[i] << " ";
    // cout << endl;

    int pr = INF;
    bool f = true;
    vector<int> v;
    // {0, n};
    for (auto itr = s.begin(); itr != s.end(); itr++)
    {
        v.push_back(*itr);
        if (*itr - pr == 1)
            f = false;
        pr = *itr;
    }
    sort(all(v));
    unsigned long long ret = 1;
    if (!f)
    {
        cout << 0 << endl;
        return 0;
    }
    else
    {
        if (v.size() == 0)
            ret = dp[n];
        else if (v.size() == 1)
            ret = (dp[v[0] - 0 - 1] * dp[n - v[0] - 1]) % MOD;
        else
        {
            // cout << v[0] - 1 << endl;
            // cout << n - v[v.size() - 1] - 1 << endl;
            ret = (ret * dp[v[0] - 0 - 1]) % MOD;
            ret = (ret * dp[n - v[v.size() - 1] - 1]) % MOD;
            for (int i = 1; i < v.size() - 1; i++)
            {
                // cout << v[i] - v[i - 1] - 2 << endl;
                ret = (ret * dp[v[i] - v[i - 1] - 2]) % MOD;
            }
        }
    }
    cout << ret << endl;
}