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.List;
  import java.util.Scanner;
  import java.util.ArrayList;
  
  public class CF1569C {
      public static long modnum = 998244353;
  
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int testCaseNum = sc.nextInt();
          for(int tc=0; tc<testCaseNum; tc++) {
              int n = sc.nextInt();
              List<Integer> a = new ArrayList<Integer>(n);
              //System.out.println("n:"+n);
              for (int i=0; i<n; i++) {
                  a.add(i, sc.nextInt());
              }
              int maxa = a.stream().max((b,c)->b>c?1:-1).get();
              long cmaxa = a.stream().filter(b->b==maxa).count();
              long k = a.stream().filter(b->b==maxa-1).count();
              long ans = 1, sub = 1;
              for (long i=1; i<=n; i++) {
                  ans = ans * i % modnum;
                  if (i != k+1) sub = sub * i % modnum;
              }
              if (cmaxa>1) {
                  System.out.println(ans);
              } else {
                  System.out.println((ans - sub + modnum) % modnum);
              }
          }
      }
  }
  
  