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.*;
  import java.util.Map.Entry;
  
  public class Div21608B {
  
      static int MOD = 998244353;
  
  	public static void main(String[] args) throws IOException {
  
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
  		
  		int t = Integer.parseInt(br.readLine());		
  	
  	    while(t --> 0){
  	        
  	        int n = Integer.parseInt(br.readLine());
  	        
  	        StringTokenizer st = new StringTokenizer(br.readLine());
  	        
  	        ArrayList<Integer> ar = new ArrayList<Integer>();
  	        
  	        for(int i = 0; i < n; i++)
  	            ar.add(Integer.parseInt(st.nextToken()));
  	        
  	        long[][] dp = new long[n+10][2];
  	        long sum = 0;
  	        
  	        for(int i = 0; i < n; i++){
  	            
  	            int val = ar.get(i);
  	            
  	            if(val > 1){
  	            
  	   	            dp[val+1][0] = (dp[val+1][0] * 2) % MOD;
      	            dp[val+1][1] = (dp[val+1][1] * 2) % MOD;
      	            
      	            dp[val-1][1] = (dp[val-1][1] * 2) % MOD;
      	            dp[val-1][1] = (dp[val-1][1] + dp[val-1][0]) % MOD;
      	            
      	            dp[val+1][0] = (dp[val+1][0] + dp[val][0]) % MOD;
  	            
  	            }else if(val == 1){
  	            
      	            dp[val+1][0] = (dp[val+1][0] * 2) % MOD;
      	            dp[val+1][1] = (dp[val+1][1] * 2) % MOD;
      	            
      	            dp[val-1][1] = (dp[val-1][1] * 2) % MOD;
      	            dp[val-1][1] = (dp[val-1][1] + dp[val-1][0]) % MOD;
      	            
      	            dp[0][1] = (dp[0][1] + 1) % MOD;
      	            
      	            dp[val+1][0] = (dp[val+1][0] + dp[val][0]) % MOD;
      	            
  	            }else{
  	                
      	            dp[val+1][0] = (dp[val+1][0] * 2) % MOD;
      	            dp[val+1][1] = (dp[val+1][1] * 2) % MOD;
      	            
      	            dp[1][0] = (dp[1][0] + 1) % MOD;
  	                
  	            }
  	            
  	        }
  	        
  	        for(int i = 0; i < n+10; i++)
  	            sum = (sum + dp[i][0] + dp[i][1]) % MOD;
  	        
  	        pw.println(sum);
  	        
  	    }
  	    
  	    pw.flush();
  
  	}
  	
  }