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 N = (long long int)(5e5 + 5);
  const long long int inf = 2e18;
  const long long int M1 = (long long int)(1e9 + 7);
  const long long int M2 = (long long int)(998244353);
  long long int n, m, a, b, c, d;
  long long int w, h;
  long long int x, y;
  long long int t;
  long long int dp[N][2];
  void solve() {
    cin >> n;
    for (long long int i = 0; i <= n + 3; i++) {
      dp[i][0] = dp[i][1] = 0;
    }
    vector<long long int> arr;
    for (long long int i = 0; i < n; i++) {
      cin >> x;
      arr.push_back(x);
    }
    long long int ans = 0;
    for (long long int i = 0; i < n; i++) {
      x = arr[i];
      if (x == 0) {
        ans += 1 + dp[0][0] + dp[2][1];
        dp[2][1] = (2 * dp[2][1]) % M2;
        dp[0][0] = (2 * dp[0][0] + 1) % M2;
      } else if (x == 1) {
        ans += dp[0][0] + dp[1][0] + dp[1][1] + dp[3][1] + 1;
        dp[3][1] = (2 * dp[3][1]) % M2;
        dp[1][1] = (2 * dp[1][1] + 1) % M2;
        dp[1][0] = (2 * dp[1][0] + dp[0][0]) % M2;
      } else {
        ans += dp[x - 2][0] + dp[x - 1][0] + dp[x][0] + dp[x][1] + dp[x + 2][1];
        dp[x + 2][1] = (2 * dp[x + 2][1]) % M2;
        dp[x][0] = (2 * dp[x][0] + dp[x - 1][0]) % M2;
        dp[x][1] = (dp[x - 2][0] + 2 * dp[x][1]) % M2;
      }
      ans = ans % M2;
    }
    cout << ans % M2 << endl;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  