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.*;
  /*
  
  
  1
  3
  0 2 1
  
  */
  
  
   
   public class D{
  	static FastReader sc=null;
  	
  	public static void main(String[] args) {
  		sc=new FastReader();
  		int t=sc.nextInt();
  		while(t-->0) {
  			int n=sc.nextInt();
  			int a[]=sc.readArray(n);
  			//0 0 .. 1 1 1 .. 2 4 2 2 4 4 2 4 ..
  			
  			//ending at has a problem of what the ending point is 
  			//so we have to write transition in terms of MEX only
  			
  			//dp1[i] -> non decreasing
  			//dp2[i] -> rest
  			
  			long dp1[]=new long[n+2],dp2[]=new long[n+2];
  			dp1[0]=1;
  			
  			for(int j:a) {
  				//011 1
  				dp1[j+1]=add(dp1[j+1],dp1[j+1]);
  				//001 2
  				dp1[j+1]=add(dp1[j+1],dp1[j]);
  				
  				//01 3 3
  				if(j>0) dp2[j-1]=add(dp2[j-1],dp2[j-1]);
  				//01 3
  				if(j>0) dp2[j-1]=add(dp2[j-1],dp1[j-1]);
  				//01 3 1
  				dp2[j+1]=add(dp2[j+1],dp2[j+1]);
  				
  				
  			}
  			
  			long ans=0;
  			for(long e:dp1)ans=add(ans,e);
  			for(long e:dp2)ans=add(ans,e);
  			ans=sub(ans,1);
  			
  			System.out.println(ans);
  			
  		}
  	}
  	static long sub(long a,long b) {
  		a-=b;
  		if(a<0)a+=mod;
  		return a;
  	}
  	static long add(long a,long b) {
  		a+=b;
  		if(a>=mod)a-=mod;
  		return a;
  	}
  	
  	static long mod=(long)998244353;
  	
  	static long mul(long a ,long b) {
  		return (a*b)%mod;
  	}
  
  	static long exp(long base,long e) {
  		if(e==0) return 1;
  		long half=exp(base,e/2);
  		if(e%2==0)return mul(half,half);
  		return mul(half,mul(half,base));
  	}
  	
  	static int[] ruffleSort(int a[]) {
  		ArrayList<Integer> al=new ArrayList<>();
  		for(int i:a)al.add(i);
  		Collections.sort(al);
  		for(int i=0;i<a.length;i++)a[i]=al.get(i);
  		return a;
  	}
  	
  	static void print(int a[]) {
  		for(int e:a) {
  			System.out.print(e+" ");
  		}
  		System.out.println();
  	}
  	
  	static class FastReader{
  		
  		StringTokenizer st=new StringTokenizer("");
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		
  		String next() {
  			while(!st.hasMoreTokens()) 
  				try {
  					st=new StringTokenizer(br.readLine());
  				}
  			   catch(IOException e){
  				   e.printStackTrace();
  			   }
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		
  		int[] readArray(int n) {
  			int a[]=new int[n];
  			for(int i=0;i<n;i++)a[i]=sc.nextInt();
  			return a;
  		}
  	}
  	
  	
  }
  