#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
constexpr int INF = 1000000001;
constexpr ll mod = 1000000007;
constexpr double PI = 3.1415926535897932;
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
inline void init() {
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
struct mint {
  ll x;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll 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() {
  init();
  int N, M;
  cin >> N >> M;
  vector<int> a(M);
  for (int i = (int)(0); i < (int)(M); ++i) cin >> a[i];
  vector<mint> dp(N + 1, 0);
  dp[0] = 1;
  for (int i = (int)(0); i < (int)(N); ++i) {
    for (int j = (int)(1); j < (int)(3); ++j) {
      if (i + j > N) continue;
      bool is = true;
      for (int k = (int)(0); k < (int)(M); ++k) {
        if (i + j == a[k]) {
          is = false;
          break;
        }
      }
      if (is) dp[i + j] += dp[i];
    }
  }
  cout << dp[N].x << "\n";
  return 0;
}
