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.ArrayList;
  import java.util.Collections;
  import java.util.HashMap;
  import java.util.PriorityQueue;
  import java.util.StringTokenizer;
  import java.io.*;
   
  public class JuryMeeting {
  	
  	private static class MyScanner {
  	    BufferedReader br;
  	    StringTokenizer st;
  
  	    public MyScanner() {
  	       br = new BufferedReader(new InputStreamReader(System.in));
  	    }
  
  	    String next() {
  	        while (st == null || !st.hasMoreElements()) {
  	            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());
  	    }
  
  	    double nextDouble() {
  	        return Double.parseDouble(next());
  	    }
  
  	    String nextLine(){
  	        String str = "";
  		  try {
  		     str = br.readLine();
  		  } catch (IOException e) {
  		     e.printStackTrace();
  		  }
  		  return str;
  	    }
  
  	 }
  	
      public static long[] fact;
      public static long m = 998244353;
      
      public static void find()
      {
      	fact[0] = 1;
      	for(int i = 1; i<=200000; i++)
      	   fact[i] = (fact[i-1] * i)%m;
      	   
      }
      
      public static long expo (long a) {
      	
      	long res = 1;
      	long b = m-2;
      	while(b>0)
      	{
      		if(b%2==1)
      			res = (res * a)%m;
      		a = (a * a)%m;
      		b/=2;
      	}
      	
      	return res;
      }
  	
  	public static long solution(int[] arr, int n)
  			
  	{
  		int max1 = Integer.MIN_VALUE;
  		
  		for(int i = 0; i<n; i++)
  		{
  			if(arr[i]>max1)
  				max1 = arr[i];
  		}
  		
          int max2 = Integer.MIN_VALUE;
  		
  		for(int i = 0; i<n; i++)
  		{
  			if(arr[i]>max2 && arr[i]<max1)
  				max2 = arr[i];
  		}
  		
  		if(max2 == Integer.MIN_VALUE)
  			return fact[n];
  		
  		int num  = 0;
  		int c = 0;
  		
  		for(int i = 0; i<n; i++)
  		{
  			if(arr[i]==max2)
  				num++;
  			if(arr[i]==max1)
  				c++;
  		}
  		
  		long ans = 0;
  		
  		if(c>1)
  			return fact[n];
  		
  		if(max1-max2>1)
  			return 0;
  		
  		for(int i = num; i<n; i++)
  			ans = (ans + ( (fact[i]*expo(fact[i-num]))%m * (fact[n-num-1]) )%m )%m;
  
  		return (fact[n] - ans + m)%m;
  			
  	}
          
  private static PrintWriter out = new PrintWriter(System.out);
  
  public static void main (String[] args)
  {
  	MyScanner s =  new MyScanner();
      fact = new long[200001];
      find();
      
      int t = s.nextInt();
      
      for(int j = 0; j<t ; j++)
      {
      	int n = s.nextInt();
      	int[] arr = new int[n];
      	for(int i =0; i<n; i++)
      		arr[i] = s.nextInt();
      	
      	out.println(solution(arr,n));
      }
      
      out.flush();
      out.close();
      
  }
  }
  