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;
  long long md = 998244353;
  long long modRecursive(long long a, long long k, long long md) {
    if (k == 0) return 1;
    long long temp = modRecursive((a * a) % md, k / 2, md) % md;
    if (k % 2) temp = (temp * a % md) % md;
    return temp;
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long t = 1;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<long long> a(n);
      map<long long, long long> mp;
      long long maxx = 0;
      for (auto &it : a) {
        cin >> it;
        maxx = max(maxx, it);
      }
      long long curr = 0, ans = 0;
      vector<long long> dp(maxx + 2), extra(maxx + 2);
      for (long long i = 0; i < n; i++) {
        if (a[i] == 0) {
          dp[a[i]] = dp[a[i]] * 2 + 1;
          extra[a[i] + 1] *= 2;
        } else if (a[i] == 1) {
          dp[a[i]] = dp[a[i]] * 2 + dp[a[i] - 1];
          extra[a[i] + 1] *= 2;
          extra[a[i] - 1] = extra[a[i] - 1] * 2 + 1;
          extra[a[i] - 1] %= md;
        } else {
          dp[a[i]] = dp[a[i]] * 2 + dp[a[i] - 1];
          extra[a[i] + 1] *= 2;
          extra[a[i] - 1] = extra[a[i] - 1] * 2 + dp[a[i] - 2];
          extra[a[i] - 1] %= md;
        }
        dp[a[i]] %= md;
        extra[a[i] + 1] %= md;
        extra[a[i]] %= md;
      }
      for (long long i = 0; i <= maxx; i++) {
        ans = (ans + dp[i] + extra[i]) % md;
      }
      cout << (ans + md) % md << endl;
    }
  }
  