description: n people gathered to hold a jury meeting of the upcoming competition, the i-th member of the jury came up with a_i tasks, which they want to share with each other.
  
  First, the jury decides on the order which they will follow while describing the tasks. Let that be a permutation p of numbers from 1 to n (an array of size n where each integer from 1 to n occurs exactly once).
  
  Then the discussion goes as follows:
  
    * If a jury member p_1 has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped. 
    * If a jury member p_2 has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped. 
    * ... 
    * If a jury member p_n has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped. 
    * If there are still members with tasks left, then the process repeats from the start. Otherwise, the discussion ends. 
  
  
  
  A permutation p is nice if none of the jury members tell two or more of their own tasks in a row. 
  
  Count the number of nice permutations. The answer may be really large, so print it modulo 998 244 353.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  The first line of the test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — number of jury members.
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of problems that the i-th member of the jury came up with.
  
  The sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case, print one integer — the number of nice permutations, taken modulo 998 244 353.
  
  Example
  
  Input
  
  
  4
  2
  1 2
  3
  5 5 5
  4
  1 3 3 7
  6
  3 4 2 1 3 3
  
  
  Output
  
  
  1
  6
  0
  540
  
  Note
  
  Explanation of the first test case from the example:
  
  There are two possible permutations, p = [1, 2] and p = [2, 1]. For p = [1, 2], the process is the following:
  
    1. the first jury member tells a task; 
    2. the second jury member tells a task; 
    3. the first jury member doesn't have any tasks left to tell, so they are skipped; 
    4. the second jury member tells a task. 
  
  
  
  So, the second jury member has told two tasks in a row (in succession), so the permutation is not nice.
  
  For p = [2, 1], the process is the following:
  
    1. the second jury member tells a task; 
    2. the first jury member tells a task; 
    3. the second jury member tells a task. 
  
  
  
  So, this permutation is nice.
solution: import java.io.BufferedReader; 
  import java.io.IOException; 
  import java.io.InputStreamReader; 
  import java.util.Scanner; 
  import java.util.StringTokenizer;
  import java.util.*;
  public class codeforcesA{
     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; 
          } 
     } 
     public static void main(String args[]){
       FastReader sc=new FastReader();
       int t=sc.nextInt();
       StringBuilder sb=new StringBuilder();
       while(t-->0){
         int n=sc.nextInt();
         int ar[]=new int[n];
         long mod=998244353;
         Map<Integer,Integer> map=new HashMap<>();
         long fact_n=1,fact_n_=1;
         int max=0,min=0;
         for(int i=0;i<n;i++){ar[i]=sc.nextInt();map.put(ar[i],map.getOrDefault(ar[i],0)+1);
            if(ar[i]>=max){min=max;max=ar[i];}else{min=Math.max(ar[i],min);}
          }
         ar[n-2]=min;ar[n-1]=max;
         for(int i=1;i<=n;i++){fact_n=(fact_n*(long)i)%mod;
            if(i!=map.get(ar[n-2])+1){fact_n_=(fact_n_*(long)i)%mod;}
          }long ans=0;
         if(ar[n-1]==ar[n-2]){
           ans=fact_n;
          }
          else{
              if(ar[n-1]-ar[n-2]==1){
               ans=(fact_n-fact_n_+mod)%mod;
              }
              else{
                  ans=0;
              }
          }
          sb.append(ans+"\n");
       }
       System.out.print(sb.toString());
      }
  }