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 {
       
       static long invfact[]=new long[200005];
       static long fact[]=new long[200005];
       static long mod= 998244353;
     public static void precompute(){
      fact[0]=fact[1]=1;
      for(int i=2;i<200005;i++)
      {
          fact[i]=fact[i-1]*(long)i;
          fact[i]%=mod;
      }
     }
    public static long modinv(long k)
  {
      return powm(k, mod-2);
  }
  public static long powm(long x, long y)
      {
          long res = 1;
          x = x % mod;
          while (y > 0) {
              if (y % 2 == 1)
                  res = (res * x) % mod;
               y = y >> 1;
              x = (x * x) % mod;
          }
          return res;
      }
    public static long nCr(int n,int r)
      {
            if (n<r)
                return 0;
          if (r == 0)
              return 1;
          return (fact[n]*modinv(fact[r])%mod*modinv(fact[n - r])%mod)%mod;
      }
       
      public static void main(String[] args) {
          
          PrintWriter out=new PrintWriter(System.out);
          Scanner sc=new Scanner(System.in);
          precompute();
          int t=sc.nextInt();
          while(t-->0) {
              int n=sc.nextInt();
              long a[]=new long[n];
              
              for(int i=0;i<n;i++) a[i]=sc.nextLong();
              sort(a);
              long total=fact[n];
              
            //  out.println(total);
              
              if(a[n-1]==a[n-2]) {
                  out.println(total);
                  continue;
              }
               if((a[n-2]+1)!=a[n-1]) {
                   out.println(0);
                 continue;
              }
             // else {
                  int countSecMax=0;
                  for(int i=0;i<n;i++) {
                      if(a[i]==a[n-2]) countSecMax++;
                  }
                  for(int i=0;i<n;i++) {
                      if(i<countSecMax) continue;
                      total=(mod+total-nCr(i,countSecMax)%mod*fact[countSecMax]%mod*fact[n-countSecMax-1]%mod)%mod;
                   
                  }
                  out.println(total);
                  
            //  }
          }
          out.flush();
          out.close();
          
          
      }
      static void sort(long a[]){
          ArrayList<Long> arr=new ArrayList<>();
          for(int i=0;i<a.length;i++)arr.add(a[i]);
          Collections.sort(arr);
          for(int i=0;i<a.length;i++)a[i]=arr.get(i);
      }
  }