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 new1{
  	
  	
  	public static long fact(int n, int m, int mod) {
  		long fact = 1;
  		for(int i = 2; i <= n; i++) {
  			if(i != m) {
  				fact = (fact * i) % mod;
  			}
  		}
  		return fact;
  	}
  
      
      public static void main(String[] args)  throws IOException{
      	FastReader s = new FastReader();
      	BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
      	int t = s.nextInt();
      	for(int i = 0; i < t; i++) {
      		int n = s.nextInt();
      		long[] arr = new long[n];
      		ArrayList<Integer> aList = new ArrayList<Integer>();
      		for(int j = 0; j < n; j++) {
      			aList.add(s.nextInt());
      		}
      		Collections.sort(aList, null);
      		for(int j = 0; j < n; j++) {
      			arr[j] = aList.get(j);
      		}
      		if(arr[n - 1] - arr[n - 2] >= 2) {
      			System.out.println(0);
      			continue;
      		}
      		long ans = -1;
      		int mod = 998244353;
      		if(arr[n - 1] == arr[n - 2]) ans = fact(n, 1, mod);
      		else {
      			int x = 1;
      			for(int j = n - 3; j >= 0; j--) {
      				if(arr[j] == arr[j + 1]) x++;
      				else break;
      			}
      			//System.out.println(fact(n, 1, mod) + " " + fact(n, x, mod));
      			ans = (fact(n, x + 1, mod) * x) % mod;
      		}
      		System.out.println(ans);
      		
      		
      		
      		output.flush();
      		
      	}
              
          
          
      }
  }
  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();
      }
   
      public 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;
  }}