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;
  const long long int mma = 1e6 + 10;
  const long long int mod = 998244353;
  vector<vector<long long int>> p;
  long long int n, m;
  bool solve(long long int x) {
    vector<bool> c(m);
    bool check = 0;
    for (int i = 0; i < n; i++) {
      long long int cf = 0;
      for (int j = 0; j < m; j++) {
        if (p[i][j] >= x) {
          cf++;
          c[j] = 1;
        }
      }
      if (cf > 1) check = 1;
    }
    if (!check) {
      return 0;
    }
    bool checkagain = 0;
    for (bool b : c)
      if (!b) return false;
    return true;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int t;
    cin >> t;
    while (t--) {
      long long int n;
      cin >> n;
      long long int dp1[n + 2], dp2[n + 2];
      fill(dp1, dp1 + n + 2, 0);
      fill(dp2, dp2 + n + 2, 0);
      dp1[0] = 1;
      long long int ans = -1;
      for (int j = 0; j < n; j++) {
        long long int x;
        cin >> x;
        dp1[x + 1] = (2 * (dp1[x + 1]));
        dp1[x + 1] = (dp1[x + 1] + dp1[x]);
        if (x > 0) {
          dp2[x - 1] = (2 * (dp2[x - 1]));
          dp2[x - 1] = (dp2[x - 1] + dp1[x - 1]);
          dp2[x - 1] %= mod;
        }
        dp2[x + 1] = (2 * (dp2[x + 1]));
        dp2[x + 1] %= mod;
        dp1[x + 1] %= mod;
      }
      for (int i = 0; i < n + 2; i++) {
        ans += dp1[i];
        ans += dp2[i];
        if (ans >= mod) ans -= mod;
      }
      cout << ans << "\n";
    }
    return 0;
  }
  