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 Main {
      public static void main(String args[])
      {
          FastReader input=new FastReader();
          PrintWriter out=new PrintWriter(System.out);
          int T=input.nextInt();
          while(T-->0)
          {
              int n=input.nextInt();
              int a[]=new int[n];
              for(int i=0;i<n;i++)
              {
                  a[i]=input.nextInt();
              }
              long dp[]=new long[n+1];
              long dp1[][]=new long[n+1][2];
              long m=998244353;
              for(int i=0;i<n;i++)
              {
                  int v=a[i];
                  long x1=0;
                  long x2=0;
                  if(v-2>=0)
                  {
                      x1=(x1+dp[v-2])%m;
                      x1=(x1+dp1[v-2][1])%m;
                  }
                  if(v+2<=n)
                  {
                      x2=(x2+dp1[v+2][0])%m;
                  }
                  x1=(x1+dp1[v][0])%m;
                  x2=(x2+dp1[v][1])%m;
                  dp1[v][0]=(dp1[v][0]+x1)%m;
                  dp1[v][1]=(dp1[v][1]+x2)%m;
                  long x=0;
                  if(v-1>=0)
                  {
                      x=(x+dp[v-1])%m;
                  }
                  x=(x+dp[v])%m;
                  dp[v]=(dp[v]+x)%m;
                  if(v==0)
                  {
                      dp[v]=(dp[v]+1)%m;
                  }
              }
  //            for(int i=0;i<=n;i++)
  //            {
  //                out.print(dp[i]+" ");
  //            }
  //            out.println();
              long ans=0;
              for(int i=0;i<=n;i++)
              {
                  ans=(ans+dp[i])%m;
                  ans=(ans+dp1[i][0])%m;
                  ans=(ans+dp1[i][1])%m;
              }
              long count=0;
              for(int i=0;i<n;i++)
              {
                  if(a[i]==1)
                  {
                      count++;
                  }
              }
              long p=power(2,count,m);
              p=(p-1)%m;
              if(p<0) p+=m;
              ans=(ans+p)%m;
              out.println(ans);
          }
          out.close();
      }
      public static long power(long a,long b,long m)
      {
          long res=1;
          while(b>0)
          {
              if(b%2!=0)
              {
                  res=(res%m*a%m)%m;
              }
              b=b/2;
              a=(a%m*a%m)%m;
          }
          return res;
      }
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer st;
          public FastReader()
          {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
          String next()
          {
              while (st == null || !st.hasMoreElements())
              {
                  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());
          }
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
          String nextLine()
          {
              String str="";
              try
              {
                  str=br.readLine();
              }
              catch (IOException e)
              {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }