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.StringTokenizer;
  
  public class CF1 {
      public static void main(String[] args) {
          FastScanner sc=new FastScanner();
          int T=sc.nextInt();
  
          for (int tt=0; tt<T; tt++) {
              long n = sc.nextInt();
              long arr[]= new long [(int) n];
              long max=Integer.MIN_VALUE;
              long countmax=0;
              for (int i =0; i<n ; i++){
                  arr[i]=sc.nextLong();
                  if (arr[i]>max) {
                      max=arr[i];
                      countmax=1;
                  }
                  else if (arr[i]==max) countmax++;
              }
              if (countmax>1) {
                  long ans=1;
                  for (int i=1; i<=n; i++){
                      ans=(ans*i)%mod;
                  }
                  System.out.println(ans);
              }
              else {
                  long max2=Integer.MIN_VALUE;
                  long countmax2=0;
                  for (int i =0; i<n ; i++){
                      if (arr[i]==max) continue;
                      else if (arr[i]>max2) {
                          max2=arr[i];
                          countmax2=1;
                      }
                      else if (arr[i]==max2) countmax2++;
                  }
                  if (max2==max-1) {
                      long wrong=1;
                      long ans=1;
                      for (int i=1; i<=n; i++){
                          ans=(ans*i)%mod;
                          if (i!=countmax2+1){
                              wrong=(wrong*i)%mod;
                          }
                      }
  
                      ans=ans-wrong;
                      System.out.println((ans+mod)%mod);
                  }
                  else System.out.println(0);
              }
          }
      }
      static long mod = 998244353;
  
  
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }
  
  