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.Arrays;
  import java.util.*;
  import java.util.Scanner;
  import java.util.StringTokenizer;
  
  
  
  public class copy{
  
  
      public static boolean checker(long[] arr,long K,long diff)
      {
          long collect=0;
          for(int i=0;i<arr.length;i++)
          {
              if(arr[i]>diff)
                  collect+=arr[i]-diff;
          }
  
          if(collect>=K)
              return true;
          else
              return false;
      }
      public static long search(long[] arr,long K,long R)
      {
          long l=0;
          long r=R;
  
          while(l<=r)
          {
              long mid=(l+r)/2;
  
              if(checker(arr,K,mid))
              {
                  if(checker(arr,K,mid+1))
                      l=mid+1;
                  else
                      return mid;
              }
              else
                  r=mid-1;
          }
          return -1;
      }
  
      static void sieveOfEratosthenes(int n,ArrayList<Integer> arr,ArrayList<Integer> arr1)
      {
          // Create a boolean array
          // "prime[0..n]" and
          // initialize all entries
          // it as true. A value in
          // prime[i] will finally be
          // false if i is Not a
          // prime, else true.
          boolean prime[] = new boolean[n + 1];
          for (int i = 0; i <= n; i++)
              prime[i] = true;
  
          for (int p = 2; p * p <= n; p++)
          {
              // If prime[p] is not changed, then it is a
              // prime
              if (prime[p] == true)
              {
  
                  // Update all multiples of p
                  for (int i = p * p; i <= n; i += p)
                      prime[i] = false;
              }
          }
  
          // Print all prime numbers
          int x=0;
          for (int i = 2; i <= n; i++)
          {
              if (prime[i] == true)
              {
  
                  arr.add(i);
  
  
              }
  
          }
          System.out.println(arr.size());
      }
  
      public static boolean check(String s,int K,char ch)
      {
          ArrayList<Integer> arr=new ArrayList<>();
          for(int i=0;i<s.length();i++)
          {
              if(s.charAt(i)==ch)
                  arr.add(i);
          }
  //        if(K==1 && ch=='z')
  //            System.out.println(arr);
          if(arr.size()==0)
              return false;
          if(arr.get(0)>=K)
              return false;
          if(s.length()-1-arr.get(arr.size()-1)>=K)
              return false;
          for(int i=1;i<arr.size();i++)
          {
              if(arr.get(i)-arr.get(i-1)>K)
                  return false;
          }
          return true;
      }
  
      public static int check(long N)
      {
          int sum=0;
          while(N>0)
          {
              sum+=N%10;
              N=N/10;
          }
          return sum;
  
      }
  
      public static long call(long N)
      {
          long l=1;
          long r=(long)Math.floor(Math.sqrt(N));
          while(l<=r)
          {
              long mid=(l+r)/2;
              double left=N/(double)(mid)-mid;
              int right=check(mid);
              System.out.println(mid+" "+left+" "+right);
              if(mid==10)
                  System.out.println(left+" "+right);
              if(left>right)
                  l=mid+1;
              else if(right>left)
                  r=mid-1;
              else
                  return mid;
          }
          return -1;
      }
  
      public static int path(ArrayList<ArrayList<Integer>> arr,int X,int parent,int[] gold,PriorityQueue<Integer> pq)
      {
          int max=0;
          ArrayList<Integer> ch=new ArrayList<>();
          for(int i:arr.get(X))
          {
              if(i!=parent)
              {
                  ch.add(path(arr,i,X,gold,pq));
              }
          }
          Collections.sort(ch,Collections.reverseOrder());
          for(int i=1;i<ch.size();i++)
              pq.add(ch.get(i));
  
  
          if(ch.size()==0)
          return gold[X];
          else
              return ch.get(0)+gold[X];
  
      }
  
      public static long fac(long N,long mod)
      {
          if(N==0)
              return 1;
          return ((N%mod)*(fac(N-1,mod)%mod))%mod;
      }
  
  
  
  
  
  
  
  
      public static void main(String[] args)throws IOException {
          Reader.init(System.in);
          BufferedWriter output=new BufferedWriter(new OutputStreamWriter(System.out));
          int T=Reader.nextInt();
          for(int m=1;m<=T;m++)
          {
              int N=Reader.nextInt();
              HashMap<Long,Long> hp=new HashMap<>();
              for(int i=0;i<N;i++)
              {
                  long X=Reader.nextLong();
                  hp.put(X,hp.getOrDefault(X,(long)0)+1);
              }
              ArrayList<Long> arr=new ArrayList<>(hp.keySet());
              long mod=998244353;
  
              Collections.sort(arr);
  
              if(hp.get(arr.get(arr.size()-1))>=2)
              {
                  output.write(fac(N,mod)+"\n");
              }
              else if(hp.get(arr.get(arr.size()-1))==1 && !hp.containsKey(arr.get(arr.size()-1)-1))
              {
                  output.write(0+"\n");
              }
              else if(hp.get(arr.get(arr.size()-1))==1)
              {
                  long freq=hp.get(arr.get(arr.size()-1)-1);
                  long select;
  
                  long fac=fac(N,mod);
                  long take=fac(freq,mod);
                  long rem=fac(N-freq-1,mod);
                  long pro=1;
                  for(long i=N;i>=N-freq;i--)
                  {
                      pro=((pro%mod)*(i%mod))%mod;
                  }
                  long pro2=1;
                  for(int i=1;i<=freq+1;i++)
                  {
                      pro2=((pro2%mod)*(i%mod))%mod;
                  }
                  long inv=power(pro2,mod-2,mod);
                  //System.out.println(take+" "+rem+" "+check);
                  select=(((((take%mod)*(pro%mod))%mod*rem%mod)%mod)*inv%mod)%mod;
                  //System.out.println(select);
                  output.write((long)(fac%mod-select%mod+mod)%mod+"\n");
  
              }
  
          }
  
  
  
  
  
          output.flush();
  
  
      }
  
      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;
      }
  
  
  
  
  
  
  }
  
  class Reader {
      static BufferedReader reader;
      static StringTokenizer tokenizer;
  
      static void init(InputStream input) {
          reader = new BufferedReader(
                  new InputStreamReader(input));
          tokenizer = new StringTokenizer("");
      }
  
      static String next() throws IOException {
          while (!tokenizer.hasMoreTokens()) {
              tokenizer = new StringTokenizer(
                      reader.readLine());
          }
          return tokenizer.nextToken();
      }
  
      static int nextInt() throws IOException {
          return Integer.parseInt(next());
      }
  
      static long nextLong() throws IOException {
          return Long.parseLong(next());
      }
  
      static double nextDouble() throws IOException {
          return Double.parseDouble(next());
      }
  }
  
  
  