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.lang.*;
  import java.io.*;
  
  public class B{
      static long MOD = 998244353;
      static long[] fact;
      static void getfact(){
          fact[0] = 1;
          for(int i=1;i<=200000;i++){
              fact[i] = (fact[i-1]*i)%MOD;
          }
      }
      static long power(long x, long y){
          long res = 1;
          x = x%MOD;
          while(y>0){
              if( y%2 == 1){
                  res = (res*x)%MOD;
              }
              y = y/2;
              x = (x*x)%MOD;
          }
          return res;
      }
      static long modInverse(long a){
          return power(a,MOD-2);
      }
      static long nCr(int n,int r){
          long sum = (long)(fact[n]%MOD);
          sum = (sum*modInverse(fact[r]))%MOD;
          sum = (sum*modInverse(fact[n-r]))%MOD;
          return sum;
      }
      public static void main(String[] args){
          FastScanner fs = new FastScanner();
          int t = fs.nextInt();
          fact = new long[200001];
          getfact();
          while(t>0){
              int n = fs.nextInt();
              long [] arr = fs.readArray(n);
              int largest = 0, second_largest = 0, freq_largest = 0, freq_second_largest = 0;
              for(int i=0;i<n;i++){
                  if(arr[i]>largest) {
                      second_largest = largest;
                      freq_second_largest = freq_largest;
                      largest = (int)arr[i];
                      freq_largest = 1;
                      continue;
                  }
                  else if(arr[i] == largest){
                      freq_largest++;
                  }else if(arr[i]> second_largest){
                      second_largest = (int)arr[i];
                      freq_second_largest = 1;
                  }else if(arr[i] == second_largest)
                      freq_second_largest++;
              }
  //            System.out.println("largest: "+largest+" second_largest: "+second_largest);
              if(freq_largest > 1)
                  System.out.println(fact[n]);
              else if(largest - second_largest > 1)
                  System.out.println(0);
              else{
                  long  curr = fact[n];
                  int  cnt = n - 1 - freq_second_largest;
                  for(int i=0;i<=cnt;i++){
                      long  sum = (nCr(cnt,i)*((fact[i]*fact[n-i-1])%MOD))%MOD;
                      curr = (curr - sum+MOD)%MOD;
                  }
                  System.out.println(curr);
              }
              t--;
          }
          return;
      }
      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());
          }
          long nextLong(){
              return Long.parseLong(next());
          }
          long[] readArray(int n){
              long[] a = new long [n];
              for(int i=0;i<n;i++)a[i] = nextLong();
              return a;
          }
      }
  }