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.util.*;
  import java.io.*;
   
  public class C {
      public static void main(String[] args) {
          FastReader sc = new FastReader();
          int t = sc.nextInt();
          for(int a=0;a<t;a++) {
              int m = sc.nextInt();
  //            sc.nextLine();
              int mod = 998244353;
  //            String[] inp = sc.nextLine().split(" ");
              Integer[] tasks = new Integer[m];
              for(int i=0;i<m;i++){
                  tasks[i] = sc.nextInt();
              }
              Collections.shuffle(Arrays.asList(tasks));
   
              Arrays.sort(tasks);
              if(tasks[m-1]-tasks[m-2]>1){
                  System.out.println(0);
              } else if (tasks[m-1]-tasks[m-2]==1) {
                  int idx = m-2;
                  int cnt=0;
                  while( idx>=0 &&tasks[idx].equals(tasks[m-2])) {
                      cnt++; idx--;
                  }
                  long res = 1;
                  for (int i = 1; i <= m; i++) {
                      if(i==(cnt+1)) continue;
                      res = (res * i) % mod;
                  }
  //                int res = modFact((m),mod);
  //                double last = ((cnt/(cnt+1)));
  //                System.out.println(last);
                  res = (res * cnt) % mod;
   
                  System.out.println(res);
              } else {
                  System.out.println(modFact(m,mod));
              }
          }
      }
   
      static int modFact(int n, int p)
      {
          long result = 1;
          for (int i = 1; i <= n; i++)
              result = (result * i) % p;
   
          return (int) result;
      }
   
      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;
          }
      }
  }
   