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;
  ifstream fin("Sisend.txt");
  long long t, n, c, cc = 998244353;
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cerr.tie(0);
    cin >> t;
    while (t--) {
      cin >> n;
      vector<long long> a = {0}, b = {0};
      for (int(i) = (0); ((i)) < ((n + 5)); ((i))++)
        a.emplace_back(0), b.emplace_back(0);
      for (int(i) = (0); ((i)) < ((n)); ((i))++) {
        cin >> c;
        if (!c)
          a[c] = (a[c] * 2 + 1) % cc;
        else
          a[c] = (a[c] * 2 + a[c - 1]) % cc;
        if (c == 1) b[c] = (b[c] * 2 + 1) % cc;
        if (c > 1) b[c] = (a[c - 2] + b[c] * 2) % cc;
        b[c + 2] = (b[c + 2] * 2) % cc;
      }
      long long vas = 0;
      for (int(i) = (0); ((i)) < ((n + 1)); ((i))++) {
        vas += a[i];
        vas %= cc;
        vas += b[i];
        vas %= cc;
      }
      cout << vas << endl;
    }
  }
  