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.util.*;
  
  public class MEX {
  
  	private static final int M = 998244353;
  	private static Scanner sc = new Scanner(System.in);
  
  	public static void main(String ...args) {
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int[] a = new int[n];
  			int[] dp1 = new int[n+1];
  			int[] dp2 = new int[n+1];
  
  			int ans = 0;
  
  			int i;
  			for (i=0; i<n; i++) {
  				a[i] = sc.nextInt();
  			}
  
  			for (i=0; i<n; i++) {
  				if (a[i] == 0 ) {
  					dp1[a[i]] = ((((dp1[a[i]] + 1) * 2) % M) -1) % M;
  					if (a[i] < n-1){
  						dp2[a[i]+2] = (dp2[a[i]+2] + dp2[a[i]+2]) % M;;
  					}
  				} else {
  					dp1[a[i]] = ((((dp1[a[i]] + dp1[a[i]]) % M) % M) + dp1[a[i]-1]) % M;
  					if (a[i] > 1) {
  						dp2[a[i]] = (((dp2[a[i]] + dp2[a[i]]) % M) + dp1[a[i]-2]) % M;
  					} else {
  						dp2[a[i]] = (((dp2[a[i]] + dp2[a[i]]) % M) + 1) % M;
  					}
  					if (a[i] < n-1) {
  						dp2[a[i]+2] = (dp2[a[i]+2] + dp2[a[i]+2]) % M;
  					}
  				}
  			}
  
  			for (i=0; i<=n; i++) {
  				ans = (ans + dp1[i]) % M;
  				ans = (ans + dp2[i]) % M;
  			}
  
  			System.out.println(ans);
  		}
  	}
  }