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>
// #include <iostream>
// #include <vector>
// #include <map>
// #include <algorithm>
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())

template<typename T>
T mod(T a, T b){
    return (a + b) % b;
}

template<typename T>
T gcd(T a, T b){
    if (b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b){
    return a * b / gcd(a,b);
}

// n を素因数分解
map<ll, ll> factor(ll n){
    map<ll, ll> mp;
    for (ll i = 2; i*i <= n; i++) {
        while (n % i == 0) {
            mp[i]++;
            n /= i;
        }
    }

    if (n > 1) {
        mp[n]++;
    }

    return mp;
}

// 文字列に "hoge" が含まれているか
// string str = "hogepiyo";
// if (str.find("hoge") != string::npos) {
//     cout << "文字列が含まれています" << endl;
// }

// char to string
// char c = 'a';
// string str = {c};

void solve(){
    ll N, M;
    cin >> N >> M;

    ll MOD = 1000000007;

    vector<ll> A(N+2, 1); // 1 --> safe, 0 --> break
    for (int i = 0; i < M; i++) {
        int k;
        cin >> k;
        A[k] = 0;
    }

    vector<ll> stair(N+2, 0);
    stair[N] = 1;
    stair[N-1] = 1;

    for (int i = N-2; i >= 0; i--) {
        ll tmp = 0;
        if (A[i+1] == 1) {
            tmp += stair[i+1];
        }
        if (A[i+2] == 1) {
            tmp += stair[i+2];
        }
        stair[i] = tmp % MOD;
    }

    cout << stair[0] << endl;
    // for (int i = 0; i < N; i++) {
    //     cout << stair[i] << endl;
    // }






    return;
}

int main(int argc, char const* argv[])
{
    solve();
    return 0;
}