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); }
  template <typename A, typename B>
  string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  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>
  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;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long long inf = 2e18;
  const double epsilon = 1e-7;
  template <typename T, typename T1>
  T amax(T &a, T1 b) {
    if (b > a) a = b;
    return a;
  }
  template <typename T, typename T1>
  T amin(T &a, T1 b) {
    if (b < a) a = b;
    return a;
  }
  inline long long gcd(long long a, long long b) {
    return (b == 0) ? a : gcd(b, a % b);
  }
  long long dx[] = {-1, 0, 1, 0};
  long long dy[] = {0, 1, 0, -1};
  void solve() {
    long long n, ans = 0;
    cin >> n;
    vector<long long> v(n);
    for (long long i = (0); i < ((n)); ++i) cin >> v[i];
    long long dp[n + 5][2];
    memset(dp, 0, sizeof(dp));
    if (v[0] <= 1) {
      if (!v[0])
        dp[1][0] = 1;
      else
        dp[0][1] = 1;
    }
    for (long long i = (1); i < (n); ++i) {
      dp[v[i] + 1][1] = (dp[v[i] + 1][1] * 2) % 998244353;
      dp[v[i] + 1][0] = (2 * dp[v[i] + 1][0]) % 998244353;
      dp[v[i] + 1][0] = (dp[v[i] + 1][0] + dp[v[i]][0]) % 998244353;
      if (v[i]) dp[v[i] - 1][1] = (2 * dp[v[i] - 1][1]) % 998244353;
      if (v[i]) dp[v[i] - 1][1] = (dp[v[i] - 1][1] + dp[v[i] - 1][0]) % 998244353;
      if (!v[i])
        dp[1][0] = (dp[1][0] + 1) % 998244353;
      else if (v[i] == 1)
        dp[0][1] = (dp[0][1] + 1) % 998244353;
    }
    for (long long i = (0); i < ((n + 1)); ++i)
      ans = (ans + dp[i][0] + dp[i][1]) % 998244353;
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long tt = 1;
    cin >> tt;
    for (long long i = 1; i <= tt; i++) solve();
  }
  