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>
  const double PI = acos(-1);
  const int N = 1e5 + 10;
  const int M = 998244353;
  long long add(long long a, long long b) { return ((a + b) % M + M) % M; }
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    vector<vector<int> > dp(n + 3, vector<int>(3));
    long long ans = 0;
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
      int val[] = {1, 1, 1};
      if (a[i] == 0) {
        val[0] = add(val[0], dp[a[i]][0]);
        val[1] = 0;
        val[2] = add(dp[a[i]][2], dp[a[i] + 2][1]);
      } else if (a[i] == 1) {
        val[0] = add(dp[a[i]][0], dp[a[i] - 1][0]);
        val[1] = add(val[1], dp[a[i]][1]);
        val[2] = add(dp[a[i]][2], dp[a[i] + 2][1]);
      } else {
        val[0] = add(dp[a[i]][0], dp[a[i] - 1][0]);
        val[1] = add(dp[a[i]][1], dp[a[i] - 2][0]);
        val[1] = add(val[1], dp[a[i] - 2][2]);
        val[2] = add(dp[a[i]][2], dp[a[i] + 2][1]);
      }
      for (int k = 0; k < 3; ++k) {
        ans = add(ans, val[k]);
        dp[a[i]][k] = add(dp[a[i]][k], val[k]);
      }
    }
    printf("%lld\n", ans);
  }
  int main() {
    int t = 1;
    scanf("%d", &t);
    while (t--) solve();
    return 0;
  }
  