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 Main {
      // For fast input output
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              try {
                  br = new BufferedReader(new FileReader("input.txt"));
                  PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
                  System.setOut(out);
              } catch (Exception e) {
                  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;
          }
      }
  
      // end of fast i/o code
      static int MOD = 998244353;
      public static void main(String[] args) {
          FastReader reader = new FastReader();
  
          int t = reader.nextInt();
          int n;
          StringBuilder sb = new StringBuilder("");
          long ans = 0;
          int i;
          int max_limit = 2*100000+1;
          long factorial[] = new long[max_limit];
          long inverse[] = new long[max_limit];
          long factorial_inv[] = new long[max_limit];
          fact(factorial, inverse, factorial_inv, max_limit);
  
          while (t-- > 0) {
              n = reader.nextInt();
              int arr[] = new int[n];
              for(i=0; i<n; i++){
                  arr[i] = reader.nextInt();
              }
  
              // System.out.println(n+"<--> "+factorial[n]);
              ans = factorial[n];
  
              
              int max = Integer.MIN_VALUE;
              for(i=0; i<n; i++){
                  max = Math.max(max, arr[i]);
              }
              
              int max_count = 0;
              for(i=0; i<n; i++){
                  if(arr[i]==max){
                      max_count++;
                  }
              }
  
              
              if(max_count>1){
                  sb.append(ans+"\n");
                  continue;
              }
  
  
              int considering = 0;
              for(i=0; i<n; i++){
                  if(arr[i]<max-1){
                      considering++;
                  }
              }
  
              if(considering==n-max_count){
                  sb.append("0\n");
                  continue;
              }
  
              long ans_temp = 0;
              for(i=0; i<=considering; i++){
  
                  ans_temp = (factorial[considering] * factorial_inv[considering-i])%MOD;
                  ans_temp = (ans_temp * factorial_inv[i])%MOD;
                  ans_temp = (ans_temp * factorial[n-i-1])%MOD;
                  ans_temp = (ans_temp * factorial[i])%MOD;
  
                  // System.out.println((n-i)+"<-,,->"+factorial_inv[n-i]+" "+ans_temp);
                  // System.out.println((((n-i)*inverse[n-i])%MOD));
                  
                  ans = (ans - ans_temp + MOD)%MOD;
                  // System.out.println(ans+"--->"+ans_temp+"\n");
              }
  
              sb.append(ans+"\n");
          }
          System.out.println(sb);
      }
  
      public static long fast_pow(long a, int b){
  
          if(b==0){
              return 1;
          }
          if(b==1){
              return a;
          }
          if((b&1)==1){
              return (a*fast_pow(a, b-1))%MOD;
          }
          return fast_pow((a*a)%MOD, b/2);
  
      }
  
      public static void fact(long factorial[], long inverse[], long factorial_inv[], int max){
  
          long current = 1;
          factorial[0] = 1;
          for(int i=1; i<max; i++){
              current = (current * i)%MOD;
              factorial[i] = current;
          }
  
  
          inverse[0] = 1;
          for(int i=1; i<max; i++){
              inverse[i] = fast_pow(i, MOD-2);
          }
          
          factorial_inv[0] = 1;
          current = 1;
          for(int i=1; i<max; i++){
              current = (current * inverse[i])%MOD;
              factorial_inv[i] = current;
          }
      }
  }