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.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class JuryMeeting{
  	private static final int MOD = 998244353;
  	public static void main(String args[]) throws Exception {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		int t = Integer.parseInt(br.readLine().trim());
  		var sb = new StringBuilder();
  		while(t-- > 0){
  			int n = Integer.parseInt(br.readLine().trim());
  			int [] input = nextIntArray(n, br);
  			sb.append(getValue(input)).append("\n");
  		}
  		br.close();
  		System.out.println(sb);
  	}
  
  	private static long getValue(int [] input){
  		int val = getMax(input);
  		int countMax = getCount(input, val);
  		int prevCount = getCount(input, val-1);
  		
  		long ret = 1, subtracter = 1;
  		for(long i = 1; i  <= input.length; i++){
  			ret = ret * i % MOD;
  			if(i != prevCount+1) subtracter = subtracter * i % MOD;
  		}
  		
  		if(countMax == 1) ret = (ret - subtracter + MOD) % MOD; 
  		
  		return ret;
  	}
  
  	private static int[] nextIntArray(int N, BufferedReader br) throws Exception {
  		StringTokenizer st = new StringTokenizer(br.readLine().trim()," ");
  		int arr[] = new int[N];
  		for(int i = 0; i < N; i++)
  		arr[i] = Integer.parseInt(st.nextToken());
  		return arr;
  	}
  	
  	private static int getCount(int [] input, int value) {
  		int count = 0;
  		for (int i : input)
  			if (i == value)
  				count++;
  
  		return count;
  	}
  
  	private static int getMax(int ...arr) {
  		int ret = Integer.MIN_VALUE;
  		for(int i : arr)
  			ret = Math.max(ret,i);
  		return ret;
  	}
  }