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;
typedef long long ll;
#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; i++)
#define reps(i, s, n) for(ll i = (s), i##_len = (n); i < i##_len; i++)
#define rrep(i, n) for(ll i = (n) - 1; i >= 0; i--)
#define rreps(i, e, n) for(ll i = (n) - 1; i >= (e); i--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())
#define len(x) ((ll)(x).length())
#define endl "\n"

struct mint {
    const int mod = 1e9 + 7;
    long long x;
    
    mint (long long x = 0) : x((x % mod + mod) % mod) {
        
    }
    
    mint& operator=(const long long a) {
        x = a % mod;
        return *this;
    }
    
    mint& operator=(const mint a) {
        x = a.x % mod;
        return *this;
    }
    
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    
    mint operator+(const mint a) const {
        mint res(*this);
        return res += a;
    }
    
    mint& operator-=(const mint a) {
        if ((x += (mod - a.x)) >= mod) x -= mod;
        return *this;
    }
    
    mint operator-(const mint a) const {
        mint res(*this);
        return res -= a;
    }
    
    mint& operator*=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }
    
    mint operator*(const mint a) const {
        mint res(*this);
        return res *= a;
    }
    
    mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }
    
    mint inv() const {
        return pow(mod - 2);
    }
    
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    
    mint operator/(const mint a) const {
        mint res(*this);
        return res /= a;
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    // ifstream in("input.txt");
    // cin.rdbuf(in.rdbuf());
    ll n, m;
    cin >> n >> m;
    set<ll> a;
    rep(i, m) {
        ll tmp;
        cin >> tmp;
        a.insert(tmp);
    }
    vector<mint> dp(n + 1, 0);
    dp[0] = 1;
    ll idx = 0;
    rep(i, n) {
        if (a.count(i + 1) > 0) continue;
        dp[i + 1] += dp[i];
        if (i > 0) dp[i + 1] += dp[i - 1];
    }
    cout << dp[n].x << endl;
    return 0;
}