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.ArrayList;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class C_1569 {
  	 static int mod=998244353;
      public static void main(String args[])
      { 
      	FastReader fs=new FastReader();
      int t=fs.nextInt();
         while(t-->0)
         {int n=fs.nextInt();
         ArrayList <Long> ar=new ArrayList<>();
         for(int i=0;i<n;i++)
      	   ar.add(fs.nextLong());
       
       long mx=Collections.max(ar);
         int mc=Collections.frequency(ar, mx);//System.out.println(mc);
         int k=Collections.frequency(ar, mx-1);//System.out.println(k);
         long sub=1,ans=1;
         ans=fact(n);
         for(long i=2;i<=n;i++)
         if (i != k + 1) sub = sub * i % mod;
       
         if(mc>=2)
      	   System.out.println(ans);
         else if(k==0)
      	   System.out.println("0");
       
         else System.out.println((ans-sub+mod)%mod);//System.out.println(fact(10000));
         }
      }
      
      static int fact(int n)
      {long result=1;
      for(long i=2;i<=n;i++)
      	result=(result*i)%mod;
      return (int) result;
      }
  }
  
  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;
      }
  }