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 Div21569C {
  
  	static long MOD = 998244353;
  	static long[] factorial = new long[200001];
  	
  	public static void main(String[] args) throws IOException {
  
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
  			
  		factorial[0] = 1;
  		for (int i = 1; i <= 200000; i++) {
  		    factorial[i] = factorial[i - 1] * i % MOD;
  		}
  		
  		int t = Integer.parseInt(br.readLine());
  		
  		while(t --> 0) {
  			
  			int n = Integer.parseInt(br.readLine());
  			StringTokenizer st = new StringTokenizer(br.readLine());
  			
  			ArrayList<Integer> ar = new ArrayList<Integer>();
  			
  			for(int i = 0; i < n; i++)
  				ar.add(Integer.parseInt(st.nextToken()));
  			
  			Collections.sort(ar, Collections.reverseOrder());
  						
  			int mc = 1;
  			int ouc = 0;
  			int mbc = 0;
  			int state = 0;
  						
  			for(int i = 1; i < n; i++) {
  				
  				if(ar.get(i).equals(ar.get(i-1))) {
  					
  					if(state == 0)
  						mc++;
  					else if(state == 1)
  						ouc++;
  					else
  						mbc++;
  					
  				}else {
  						
  					if(ar.get(i) + 1 == ar.get(i-1))
  						state++;
  					else
  						state += 2;
  					
  					if(state == 1)
  						ouc++;
  					else
  						mbc++;
  					
  				}
  				
  			}
  			
  			if(mc >= 2 || n == 1) {
  				
  				pw.println(factorial[n]);
  				
  			}else {
  				
  				if(ouc == 0) {
  					
  					pw.println(0);
  					
  				}else {
  					
  					long val = factorial[n];
  															
  					for(int i = 0; i <= mbc; i++) {
  						
  						val -= (binomial_coefficient(mbc, i) * (factorial[ouc + mbc - i] * factorial[i] % MOD)) % MOD;
  						while(val < 0)
  							val += MOD;
  												
  					}
  					
  					pw.println(val);
  					
  				}
  				
  			}
  									
  		}
  		
  		pw.close();
  
  	}
  	
  	static long binpow(long a, long b, long m) {
  	    a %= m;
  	    long res = 1;
  	    while (b > 0) {
  	        if ((b & 1) == 1)
  	            res = res * a % m;
  	        a = a * a % m;
  	        b >>= 1;
  	    }
  	    return res;
  	}
  	
  	static long binomial_coefficient(int n, int k) {
  		if(k == 0)
  			return 1;
  	    return factorial[n] * binpow(factorial[k] * factorial[n - k] % MOD, MOD-2, MOD) % MOD;
  	}
  
  }
  