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.StringTokenizer;
  import java.io.InputStreamReader;
  import java.lang.reflect.Array;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.*;
  import java.math.BigInteger;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.ArrayList;
  
  public class p3
  {
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader()
  		{
  				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()); }
  
  		byte nextByte() { return Byte.parseByte(next()); }
  
  		short nextShort() { return Short.parseShort(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 void main(String[] args)throws Exception
  	{
  		FastReader fr=new FastReader();
  		BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
  
  		short t=fr.nextShort();
  		while(t-->0)
  		{
  			int n=fr.nextInt();
  			int a[]=new int[n];
  			for(int i=-1;++i<n;)
  				a[i]=fr.nextInt();
  
  			//3 4 2 1 3 3
  			//2 3 1 - 2 2
  			//1 2 - - 1 1
  			//
  
  			int max=0;
  			int maxc=0;
  			int max2=0;
  			int max2c=0;
  			int mod=998244353;
  			for(int i=-1;++i<n;)
  			{
  				if(a[i]>max)
  				{
  					max=a[i];
  					maxc=0;
  				}
  				if(a[i]==max)
  					maxc++;				
  			}
  
  			if(maxc>1)
  			{
  				long ans=1;
  				for(int i=0;++i<=n;)
  				{
  					ans*=i;
  					ans%=mod;
  				}
  				System.out.println(ans);
  			}
  			else
  			{
  				for(int i=-1;++i<n;)
  				{
  					if(a[i]>max2 && a[i]<max)
  					{
  						max2=a[i];
  						max2c=0;
  					}
  					if(a[i]==max2)
  						max2c++;
  				}
  
  				if(max-max2>1)
  				{
  					System.out.println(0);
  				}
  				else
  				{
  					long ans=1;
  					for(int i=0;++i<=n;)
  					{
  						if(i==max2c+1)
  						{
  							ans*=max2c;
  							ans%=mod;
  							continue;
  						}
  						ans*=i;
  						ans%=mod;
  					}
  					System.out.println(ans);
  				}
  			}
  
  		}
  	}
  }