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.*;
  import java.util.*;
  
  public class C {
      static FastScanner sc;
      static PrintWriter pw;
  
      static long MOD = 998244353; 
  
      public static void main(String[] args) throws Exception{
          sc = new FastScanner();
          pw = new PrintWriter(System.out);
          
          try{
  
              int T = sc.ni();
              while(T-->0){
                  int n = sc.ni();
                  long a[] = new long[n];
                  long max1 = Long.MIN_VALUE;
                  long max2 = Long.MIN_VALUE;
                  Map<Long,Integer> mp = new HashMap<>();
                  for(int i=0; i<n; i++){
                      a[i] = sc.nl();
                      mp.put(a[i], mp.getOrDefault(a[i], 0)+1);
                      if(a[i]>max1){
                          max2 = max1;
                          max1 = a[i];
                      }else if(a[i]>max2){
                          max2 = a[i];
                      }
                  }
                  if(mp.get(max1)>=2){
                      long ans = factoial(n);
                      pw.println(ans);
                  }else if(max2<=(max1-2)){
                      pw.println(0);
                  }else{
                      int k = mp.get(max2);
                      long ans = 1;
                      long s = 1;
                      for(long i=1; i<=n; i++){
                          ans = ans * i %MOD;
                          if(i!=k+1){
                              s = s * i % MOD;
                          }
                      }
                      ans = (ans - s + MOD) % MOD;
                      pw.println(ans);
                  }
              }
  
          }catch(Exception e){
              return;
          }
  
          pw.close();
      }
  
      static long factoial(int n){
          if (n >= MOD)
              return 0;
          long result = 1;
          for (long i = 1; i <= n; i++)
              result = ((result%MOD) * (i%MOD))%MOD;
       
          return result;
      }
  
      static long nCr(int n, int r){
  		if (r > n - r)
  			r = n - r;
  		long C[] = new long[r + 1];
  		C[0] = 1;
  		for (int i = 1; i <= n; i++) {
  			for (int j = Math.min(i, r); j > 0; j--)
  				C[j] = ((C[j]%MOD) + (C[j - 1]%MOD)) % MOD;
  		}
  		return C[r];
  	}
  
      static class FastScanner {
          BufferedReader br;
          StringTokenizer st;
   
          public FastScanner() {
              br = new BufferedReader(new InputStreamReader(System.in), 32768);
              st = null;
          }
   
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
   
          int ni() {
              return Integer.parseInt(next());
          }
  
          int[] intArray(int N) {
              int[] ret = new int[N];
              for (int i = 0; i < N; i++)
                  ret[i] = ni();
              return ret;
          }
   
          long nl() {
              return Long.parseLong(next());
          }
  
          long[] longArray(int N) {
              long[] ret = new long[N];
              for (int i = 0; i < N; i++)
                  ret[i] = nl();
              return ret;
          }
  
          double nd() {
              return Double.parseDouble(next());
          }
   
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
  }