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.*;
  //solve harder problems
  //upsolve
  //use gitgud
  //learn from everything
  //USACO
  public class Contest1569C
  {
  	static class InputReader {
  		BufferedReader reader;
  		StringTokenizer tokenizer;
  		public InputReader(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream), 32768);
  			tokenizer = null;
  		}
  		String next() { // reads in the next string
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				try {
  					tokenizer = new StringTokenizer(reader.readLine());
  				} catch (IOException e) {
  					throw new RuntimeException(e);
  				}
  			}
  			return tokenizer.nextToken();
  		}
  		public int nextInt() { // reads in the next int
  			return Integer.parseInt(next());
  		}
  		public long nextLong() { // reads in the next long
  			return Long.parseLong(next());
  		}
  		public double nextDouble() { // reads in the next double
  			return Double.parseDouble(next());
  			}
  		}
  	static InputReader r = new InputReader(System.in);
  	static PrintWriter pw = new PrintWriter(System.out);
  	static long m = 998244353;
  	public static void main(String[] args)
  	{
  		//LOOK FOR INT OVERFLOW
  		//LOOK FOR SCOPE ERROR
  		//Add negatives
  		//SORT BOTH ARRAYS
  		//CHECK THE LAST ONE THAT IS OUT LOOP
  		//SORT BY BOTH VALUES
  		//DON'T ADD STRINGS,USE STRING BUILDER
  		int t = r.nextInt();
  		while (t > 0)
  		{
  			t--;
  			int n = r.nextInt();
  			int[] a = new int[n];
  			int max = 0;
  			for (int i = 0; i < n; i ++)
  			{
  				a[i] = r.nextInt();
  				max = Math.max(a[i], max);
  			}
  			int countM = 0;
  			int countm = 0;
  			for (int i = 0; i < n; i ++)
  			{
  				if (a[i] == max)
  				{
  					countM ++;
  				}
  				else if (a[i] == max - 1)
  				{
  					countm ++;
  				}
  			}
  			if (countM > 1)
  			{
  				pw.println(factorial(n));
  			}
  			else if (countm == 0)
  			{
  				pw.println(0);
  			}
  			else
  			{
  				pw.println((factorial(n)-((factorial(n)*(binpow(countm+1,m-2)%m))%m)+m)%m);
  			}
  		}
  		
  		pw.close();
  	}
  	public static long factorial(long x)
  	{
  		if (x == 1 || x == 0)
  		{
  			return 1;
  		}
  		return ((x%m) * (factorial(x-1)%m))%m;
  	}
  	public static long binpow(long x, long n) {
  		assert(n >= 0);
  		x %= m;  // note: m * m must be less than 2^63 to avoid ll overflow
  		long res = 1;
  		while (n > 0) {
  			if (n % 2 == 1)  // if n is odd
  				res = res * x % m;
  			x = x * x % m;
  			n /= 2;  // divide by two
  		}
  		return res;
  	}
  }