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 d {
      public static int mod = 998244353;
  
      public static void main(String[] args) throws Exception {
          BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter out = new PrintWriter(System.out);
          StringBuilder b = new StringBuilder();
          int numcases = Integer.parseInt(in.readLine());
          for (int casenum = 0; casenum < numcases; casenum++) {
              int n = Integer.parseInt(in.readLine());
              int[] arr = new int[n];
              StringTokenizer tokenizer = new StringTokenizer(in.readLine());
              for (int i = 0; i < n; i++) {
                  arr[i] = Integer.parseInt(tokenizer.nextToken());
              }
              int[][] dp = new int[n + 5][2];
              for (int i = 0; i < n; i++) {
                  if (arr[i] > 0) {
                      add(dp[arr[i] - 1], 1, dp[arr[i] - 1][1]);
                      add(dp[arr[i] - 1], 1, dp[arr[i] - 1][0]);
                  }
                  add(dp[arr[i] + 1], 0, dp[arr[i] + 1][0]);
                  add(dp[arr[i] + 1], 1, dp[arr[i] + 1][1]);
                  add(dp[arr[i] + 1], 0, dp[arr[i]][0]); // depends on if arr[i] + 1 is already contained
                  if (arr[i] == 1) {
                      add(dp[0], 1, 1);
                  }
                  if (arr[i] == 0) {
                      add(dp[1], 0, 1);
                  }
              }
              // System.out.println(Arrays.toString(dp));
              int sum = 0;
              for (int i = 0; i < dp.length; i++) {
                  sum += dp[i][0];
                  sum %= mod;
                  sum += dp[i][1];
                  sum %= mod;
              }
              b.append(sum + "\n");
          }
          System.out.print(b);
          in.close();
          out.close();
      }
  
      public static void add(int[] arr, int index, int val) {
          arr[index] += val;
          if (arr[index] >= mod) {
              arr[index] %= mod;
          }
      }
  }