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;
  vector<string> split(const string& s, char c) {
    vector<string> v;
    stringstream ss(s);
    string x;
    while (getline(ss, x, c)) v.emplace_back(x);
    return move(v);
  }
  template <typename T, typename... Args>
  inline string arrStr(T arr, int n) {
    stringstream s;
    s << "[";
    for (int i = 0; i < n - 1; i++) s << arr[i] << ",";
    s << arr[n - 1] << "]";
    return s.str();
  }
  inline void __evars_begin(int line) { cerr << "#" << line << ": "; }
  template <typename T>
  inline void __evars_out_var(vector<T> val) {
    cerr << arrStr(val, val.size());
  }
  template <typename T>
  inline void __evars_out_var(T* val) {
    cerr << arrStr(val, 10);
  }
  template <typename T>
  inline void __evars_out_var(T val) {
    cerr << val;
  }
  inline void __evars(vector<string>::iterator it) { cerr << endl; }
  template <typename T, typename... Args>
  inline void __evars(vector<string>::iterator it, T a, Args... args) {
    cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
    __evars_out_var(a);
    cerr << "; ";
    __evars(++it, args...);
  }
  const long long N = 2e5 + 15;
  const long long sz = (1 << 20);
  const long long mod = 1e9 + 7;
  const long long mod2 = 998244353;
  void boomer() {
    long long n;
    cin >> n;
    vector<long long> v(n + 1);
    vector<long long> a;
    for (long long i = 1; i <= n; ++i) {
      cin >> v[i];
      a.push_back(v[i]);
    }
    sort(a.begin(), a.end());
    long long mex = 0;
    for (long long x : a) {
      if (x == mex) mex++;
    }
    vector<long long> dp1(mex + 10), dp2(mex + 10);
    for (long long i = 1; i <= n; ++i) {
      if (v[i]) {
        long long x = v[i] - 1;
        dp2[x] = (dp2[x] + dp2[x] + dp1[x]) % mod2;
      }
      long long x = v[i] + 1;
      dp1[x] = (dp1[x] + dp1[x]) % mod2;
      dp2[x] = (dp2[x] + dp2[x]) % mod2;
      dp1[x] = (dp1[x] + dp1[x - 1]) % mod2;
      if (v[i] == 0) dp1[1]++;
      if (v[i] == 1) dp2[0]++;
    }
    long long ans = 0;
    for (long long i = 0; i <= mex; ++i) ans = (ans + dp1[i] + dp2[i]) % mod2;
    cout << ans;
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long tc = 1;
    cin >> tc;
    for (long long i = 1; i <= tc; ++i) {
      boomer();
      if (i < tc) cout << "\n";
    }
    return 0;
  }
  