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 N, A[1000001], Dp[1000001], E[1000001];
  inline long long Read() {
    register long long Return(0), Flag(0);
    register char C(getchar());
    for (; !isdigit(C); C = getchar()) Flag ^= !(C ^ 45);
    for (; isdigit(C); C = getchar())
      Return = (Return << 1) + (Return << 3) + (C ^ '0');
    if (Flag) Return = -Return;
    return Return;
  }
  inline void Write(long long X) {
    if (X < 0) X = -X, putchar('-');
    if (X > 9) Write(X / 10);
    putchar(X % 10 ^ 48);
  }
  signed main(void) {
    ios::sync_with_stdio(false);
    register long long Case(Read());
    while (Case--) {
      register long long i;
      N = Read();
      for (i = 1; i <= N + 5; i++) Dp[i] = E[i] = 0;
      Dp[0] = 1;
      for (i = 1; i <= N; i++) A[i] = Read();
      register long long Ret(0);
      for (i = 1; i <= N; i++) {
        Ret += E[A[i]] + E[A[i] + 2];
        if (A[i]) Ret += Dp[A[i] - 1];
        Ret %= 998244353;
        E[A[i]] = 2 * E[A[i]];
        E[A[i] + 2] = 2 * E[A[i] + 2];
        if (A[i]) E[A[i]] += Dp[A[i] - 1];
        E[A[i]] %= 998244353;
        E[A[i] + 2] %= 998244353;
        if (!A[i])
          Ret += Dp[0] + Dp[1], Ret %= 998244353;
        else
          Ret += Dp[A[i] + 1] + Dp[A[i]], Ret %= 998244353;
        Dp[A[i] + 1] = Dp[A[i] + 1] * 2 + Dp[A[i]], Dp[A[i] + 1] %= 998244353;
      }
      Write(Ret), putchar('\n');
    }
    return 0;
  }
  