description: Let's call a sequence of integers x_1, x_2, ..., x_k MEX-correct if for all i (1 ≤ i ≤ k) |x_i - \operatorname{MEX}(x_1, x_2, ..., x_i)| ≤ 1 holds. Where \operatorname{MEX}(x_1, ..., x_k) is the minimum non-negative integer that doesn't belong to the set x_1, ..., x_k. For example, \operatorname{MEX}(1, 0, 1, 3) = 2 and \operatorname{MEX}(2, 1, 5) = 0.
  
  You are given an array a consisting of n non-negative integers. Calculate the number of non-empty MEX-correct subsequences of a given array. The number of subsequences can be very large, so print it modulo 998244353. 
  
  Note: a subsequence of an array a is a sequence [a_{i_1}, a_{i_2}, ..., a_{i_m}] meeting the constraints 1 ≤ i_1 < i_2 < ... < i_m ≤ n. If two different ways to choose the sequence of indices [i_1, i_2, ..., i_m] yield the same subsequence, the resulting subsequence should be counted twice (i. e. two subsequences are different if their sequences of indices [i_1, i_2, ..., i_m] are not the same).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 5 ⋅ 10^5).
  
  The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ n).
  
  The sum of n over all test cases doesn't exceed 5 ⋅ 10^5.
  
  Output
  
  For each test case, print a single integer — the number of non-empty MEX-correct subsequences of a given array, taken modulo 998244353.
  
  Example
  
  Input
  
  
  4
  3
  0 2 1
  2
  1 0
  5
  0 0 0 0 0
  4
  0 1 2 3
  
  
  Output
  
  
  4
  2
  31
  7
  
  Note
  
  In the first example, the valid subsequences are [0], [1], [0,1] and [0,2].
  
  In the second example, the valid subsequences are [0] and [1].
  
  In the third example, any non-empty subsequence is valid. 
solution: #include <bits/stdc++.h>
  using namespace std;
  string to_string(char c) { return string(1, c); }
  string to_string(string s) { return s; }
  string to_string(const char* s) { return string(s); }
  string to_string(bool b) { return (b ? "true" : "false"); }
  string to_string(vector<bool> v) {
    string res;
    for (int i = 0; i < (int)v.size(); i++) res += char('0' + v[i]);
    return res;
  }
  template <size_t N>
  string to_string(bitset<N> v) {
    string res = "";
    for (size_t i = 0; i < N; i++) {
      res += static_cast<char>('0' + v[i]);
    }
    return res;
  }
  template <typename A, typename B>
  string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  template <typename A, typename B, typename C>
  string to_string(tuple<A, B, C> p) {
    return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
           to_string(get<2>(p)) + ")";
  }
  template <typename A, typename B, typename C, typename D>
  string to_string(tuple<A, B, C, D> p) {
    return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
           to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
  }
  template <typename A>
  string to_string(A v) {
    bool first = true;
    string res = "{";
    for (const auto& x : v) {
      if (!first) {
        res += ", ";
      }
      first = false;
      res += to_string(x);
    }
    res += "}";
    return res;
  }
  void debug_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cout << " " << to_string(H);
    debug_out(T...);
  }
  mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  using V = vector<T>;
  template <typename T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <typename T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  const long long dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
  using ll = long long;
  using db = long double;
  using vd = vector<db>;
  using vs = vector<string>;
  using pi = pair<long long, long long>;
  using vi = vector<long long>;
  using vb = vector<bool>;
  using vl = vector<ll>;
  using vpi = vector<pi>;
  template <class A>
  void read(V<A>& v);
  template <class A, size_t S>
  void read(array<A, S>& a);
  template <class T>
  void read(T& x) {
    cin >> x;
  }
  void read(double& d) {
    string t;
    read(t);
    d = stod(t);
  }
  void read(long double& d) {
    string t;
    read(t);
    d = stold(t);
  }
  template <class H, class... T>
  void read(H& h, T&... t) {
    read(h);
    read(t...);
  }
  template <class A>
  void read(V<A>& x) {
    for (auto& a : x) read(a);
  }
  template <class A, size_t S>
  void read(array<A, S>& x) {
    for (auto& a : x) read(a);
  }
  const long long mod = 998244353;
  const ll INF = 1e18;
  const db PI = acos((db)-1);
  constexpr long long pct(long long x) { return __builtin_popcount(x); }
  constexpr long long bits(long long x) {
    return x == 0 ? 0 : 31 - __builtin_clz(x);
  }
  constexpr long long p2(long long x) { return 1 << x; }
  constexpr long long msk2(long long x) { return p2(x) - 1; }
  template <typename T>
  T binpow(T a, T b) {
    T ans = 1;
    while (b) {
      if (b & 1) {
        ans = 1LL * ans * a % mod;
      }
      a = 1LL * a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  void solve() {
    long long n;
    cin >> n;
    vi a(n);
    read(a);
    V<vi> dp(n + 100, vi(3, 0));
    for (long long i = (0); i < (n); ++i) {
      long long x = a[i];
      dp[x][2] = (2 * dp[x][2] + (x - 1 >= 0 ? dp[x - 1][2] : 0)) % mod;
      dp[x][0] = (2 * dp[x][0] + (x >= 2 ? dp[x - 2][1] : 0)) % mod;
      dp[x][1] =
          (2 * dp[x][1] + (x >= 1 ? dp[x - 1][2] : 0) + dp[x + 2][0]) % mod;
      if (!x) {
        dp[x][1] = (dp[x][1] + 1) % mod;
        dp[0][2] = (dp[0][2] + 1) % mod;
      }
    }
    long long ans = 0;
    for (long long i = (0); i < (n + 1); ++i)
      ans = (ans + dp[i][0] + dp[i][1]) % mod;
    long long x = count((a).begin(), (a).end(), 1);
    x = (binpow<long long>(2, x) - 1 + mod) % mod;
    ans = (ans + x) % mod;
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long tt = 1;
    cin >> tt;
    for (long long ii = 1; ii <= tt; ii++) solve();
  }
  