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: import java.io.*;
  import java.util.*;
  public class Ishu
  {
      static final int MOD = 998244353;
      static Scanner scan = new Scanner(System.in);
      static BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out)); 
      static void tc() throws Exception
      {
      int n = scan.nextInt();
      int[] a = new int[n];
      int i, j;
      
      for(i=0;i<n;++i)
          a[i] = scan.nextInt();
          
      long ans = 0l;
      long[] l = new long[n + 1];
      long[] r = new long[n + 1];
      long[] cnt = new long[n + 1];
      long zero = 1;  // l[-1]
      long one = 0;   // r[-1]
      for(i=0;i<n;++i)
          {
          int ce = a[i];
          if(ce == 0)
              {
              ans = (ans + ((zero + l[ce]) % MOD + r[ce]) % MOD) % MOD;
              
              l[ce] = (l[ce] + (l[ce] + zero) % MOD) % MOD;
              r[ce] = (r[ce] + r[ce]) % MOD;
              }
          else if(ce == 1)
              {
              ans = (ans + ((((zero + one) % MOD + l[ce - 1]) % MOD + l[ce]) % MOD + r[ce]) % MOD) % MOD;
              
              one = (one + 1 + one) % MOD;
              l[ce] = (l[ce] + (l[ce] + l[ce - 1]) % MOD) % MOD;
              r[ce] = (r[ce] + r[ce]) % MOD;
              }   
          else
              {   
              ans = (ans + ((((l[ce - 2] + r[ce - 2]) % MOD + l[ce - 1]) % MOD + l[ce]) % MOD + r[ce]) % MOD) % MOD;
              
              r[ce - 2] = (r[ce - 2] + r[ce - 2] + l[ce - 2]) % MOD;
              l[ce] = (l[ce] + (l[ce] + l[ce - 1]) % MOD) % MOD;
              r[ce] = (r[ce] + r[ce]) % MOD;
              }
         // System.out.println(i + " " + ans + " " + one + " " + zero);
          }
  
      output.write(ans + "\n");
      output.flush();
      }
      
      public static void main(String[] args) throws Exception
      {
      int t = 1;
      t = scan.nextInt();
      while(t-- > 0)
          tc();
      }
  }
  /*
  1 = [1]
  1 = []
  2
  5
  zero = 1;
  one = 1;
  ans = 1 + 1
  
  */
  
  