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 Codeforces
  {
      public static void main(String args[])throws Exception
      {
          BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
          StringBuilder sb=new StringBuilder();
          int i,N=200000;
          long f[]=new long[N+1],in[]=new long[N+1],M=998244353;
          f[0]=1;
          for(i=1;i<=N;i++) f[i]=f[i-1]*i%M;
          in[N]=power(f[N],M-2,M);
          for(i=N-1;i>=0;i--) in[i]=in[i+1]*(i+1)%M;
  
          int t=Integer.parseInt(bu.readLine());
          while(t-->0)
          {
              int n=Integer.parseInt(bu.readLine());
              int a[]=new int[n];
              String s[]=bu.readLine().split(" ");
  
              PriorityQueue<Integer> pq=new PriorityQueue<>();
              for(i=0;i<n;i++)
              {
                  a[i]=Integer.parseInt(s[i]);
                  pq.add(a[i]);
              }
              for(i=0;i<n;i++) a[i]=pq.poll();
  
              long ans=0;
              if(a[n-1]-a[n-2]>1) ans=0;
              else if(a[n-1]==a[n-2]) ans=f[n];
              else
              {
                  int secl=0;
                  for(i=0;i<n;i++)
                  if(a[i]==a[n-2]) secl++;
  
                  secl=n-(secl+1);
                  for(i=0;i<n;i++)
                  if(secl>=n-1-i)
                  ans=(ans+ncr(f,in,secl,n-1-i)*f[n-1-i]%M*f[i]%M)%M;
                  ans=(f[n]-ans+M)%M;
              }
              sb.append(ans+"\n");
          }
          System.out.print(sb);
      }
  
      static long ncr(long f[],long in[],int n,int r)
      {
          if(n<r) return 0;
          long M=998244353;
          return f[n]*in[r]%M*in[n-r]%M;
      }
  
      static long power(long a,long b,long M)
      {
          long res=1;
          while(b!=0)
          {
              if(b%2==1) res=res*a%M;
              b>>=1;
              a=a*a%M;
          }
          return res;
      }
  }