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.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class JuryMeeting {
      final static int MOD = 998244353;
  
      public static long qpow(long base, long exp){
          long result = 1;
  
          while(exp > 0){
              if ((exp&1) == 1) result = result*base%MOD;
              base = base*base%MOD;
              exp >>= 1;
          }
          return  result%MOD;
      }
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter pw =new PrintWriter(System.out);
  
          StringBuilder sb = new StringBuilder();
          int t = Integer.parseInt(br.readLine());
  
          for (int i = 0; i < t; i++) {
              int n = Integer.parseInt(br.readLine());
              //total permutaions - not nice permutations
              //it is not going to be nice only when the largest (only one largest) is after all the other second largest
  
              StringTokenizer st = new StringTokenizer(br.readLine());
              int[] tasks = new int[n];
              int max = 0;
              int maxCount = 0;
              for (int j = 0; j <n; j++) {
                  int num = Integer.parseInt(st.nextToken());
                  tasks[j] = num;
                  if (num > max){
                      max = num;
                      maxCount = 1;
                  }else if (num == max){
                      maxCount ++;
                  }
              }
              int secCount = 0;
              for (int j = 0; j <n; j++) {
                  int num = tasks[j];
                  if (num == max - 1)
                      secCount ++;
              }
              long ans = 1;
              long sub  =1;
  
              for(int j = 1; j <= n ;j++){
                  ans =  ans * j %MOD;
                  if (j != secCount + 1) sub = sub *j%MOD;
              }
              if (maxCount > 1) sb.append(ans + "\n");
  //            else sb.append((ans - sub + MOD)%MOD + "\n");
              else sb.append((ans - ans*qpow(secCount+1, MOD-2)%MOD + MOD)%MOD + "\n");
  
  
          }
          br.close();
          pw.print(sb);
          pw.close();
      }
  }
  