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 mod = 998244353;
  long long binp(long long a, long long p) {
    if (p == 0) return 1;
    if (p & 1) {
      return ((long long)a * binp(a, p - 1)) % mod;
    } else {
      long long t = binp(a, p / 2);
      return (t * t) % mod;
    }
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    vector<vector<long long>> pos(n + 1);
    for (long long i = 0; i < n; i++) pos[a[i]].push_back(i);
    vector<long long> dpprev(n, 0);
    vector<long long> dpcur(n, 0);
    vector<long long> dpans(n, 0);
    long long cnt0 = 0;
    long long res = 0;
    long long cnt1 = 0;
    for (long long i = 0; i < n; i++)
      if (a[i] == 1) cnt1++;
    res = binp(2, cnt1) - 1;
    long long cnt02 = 0;
    for (long long i = 0; i < n; i++)
      if (a[i] == 0 || a[i] == 2) cnt02++;
    for (long long i = 0; i < n; i++) {
      if (a[i] != 0 && a[i] != 2) continue;
      if (a[i] == 0) res = (res + binp(2, cnt02 - 1)) % mod;
      cnt02--;
    }
    for (long long i = 0; i < n; i++) {
      if (a[i]) continue;
      dpprev[i] = binp(2, cnt0);
      cnt0++;
    }
    for (long long i = 1; i <= n; i++) {
      long long id = 0;
      long long idn = 0;
      for (auto e : pos[i]) dpcur[e] = dpans[e] = 0;
      if (pos[i].size()) {
        while (id < pos[i - 1].size() && pos[i - 1][id] < pos[i][0]) {
          dpcur[pos[i][0]] = (dpcur[pos[i][0]] + dpprev[pos[i - 1][id]]) % mod;
          dpans[pos[i][0]] = (dpans[pos[i][0]] + dpprev[pos[i - 1][id]]) % mod;
          id++;
        }
        long long cntcurnext = 0;
        if (i + 2 <= n)
          cntcurnext = pos[i].size() + pos[i + 2].size();
        else
          cntcurnext = pos[i].size();
        while (i + 2 <= n && idn < pos[i + 2].size() &&
               pos[i + 2][idn] < pos[i][0]) {
          cntcurnext--;
          idn++;
        }
        res = (res +
               ((long long)dpcur[pos[i][0]] * binp(2, cntcurnext - 1)) % mod) %
              mod;
        cntcurnext--;
        for (long long j = 1; j < pos[i].size(); j++) {
          dpans[pos[i][j]] = dpans[pos[i][j - 1]];
          dpcur[pos[i][j]] = (2 * dpcur[pos[i][j - 1]]) % mod;
          while (i + 2 <= n && idn < pos[i + 2].size() &&
                 pos[i + 2][idn] < pos[i][j]) {
            cntcurnext--;
            idn++;
          }
          while (id < pos[i - 1].size() && pos[i - 1][id] < pos[i][j]) {
            dpcur[pos[i][j]] = (dpcur[pos[i][j]] + dpprev[pos[i - 1][id]]) % mod;
            dpans[pos[i][j]] = (dpans[pos[i][j]] + dpprev[pos[i - 1][id]]) % mod;
            id++;
          }
          res = (res +
                 ((long long)dpans[pos[i][j]] * binp(2, cntcurnext - 1)) % mod) %
                mod;
          cntcurnext--;
        }
      }
      for (auto e : pos[i]) dpprev[e] = dpcur[e];
    }
    cout << res << "\n";
    return;
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long test_cnt = 1;
    cin >> test_cnt;
    for (long long test = 1; test <= test_cnt; test++) {
      solve();
    }
    return 0;
  }
  