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.*;
  import java.io.*;
  
  public class C_1569 {
  	
  	public static int pow(int a, int b, int m) {
  		int res = 1;
  		while(b > 0) {
  			if((b & 1) == 1)
  				res = (int)((1l * res * a) % m);
  			a = (int)((1l * a * a) % m);
  			b >>= 1;
  		}
  		return res;
  	}
  	
  	public static void main(String[] args) throws Exception {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		
  		int mod = 998244353;
  		int[] fac = new int[(int)2e5 + 1];
  		fac[0] = 1;
  		for(int i = 1; i < fac.length; i++)
  			fac[i] = (int)((1l * fac[i - 1] * i) % mod);
  		
  		int t = sc.nextInt();
  		while(t-->0) {
  			int n = sc.nextInt();
  			int[] array = sc.nextIntArray(n);
  			TreeMap<Integer, Integer> map = new TreeMap<>();
  			for(int i = 0; i < n; i++)
  				map.put(array[i], map.getOrDefault(array[i], 0) + 1);
  			
  			if(map.get(map.lastKey()) > 1) {
  				pw.println(fac[n]);
  			} else {
  				if(map.lowerKey(map.lastKey()) != map.lastKey() - 1)
  					pw.println(0);
  				else {
  					int b = map.pollLastEntry().getValue(), a = map.pollLastEntry().getValue();
  					int all = fac[n];
  					
  					int ans = (int)((1l * fac[n]) % mod);
  					int div = (int)((1l * fac[a + b] * fac[n - a - b]) % mod);
  					div = pow(div, mod - 2, mod);
  					ans = (int)((1l * ans * div) % mod);
  					ans = (int)((1l * ans * fac[a + b]) % mod);
  					
  					int ans2 = (int)((1l * fac[n]) % mod);
  					int div2 = (int)((1l * fac[a + b] * fac[n - a - b]) % mod);
  					div2 = pow(div2, mod - 2, mod);
  					ans2 = (int)((1l * ans2 * div2) % mod);
  					ans2 = (int)((1l * ans2 * fac[a]) % mod);
  					
  					ans = (int)((1l * ans - ans2) % mod);
  					ans = ((ans % mod) + mod) % mod;
  					
  					ans = (int)((1l * ans * fac[n - a - b]) % mod);
  					
  					//all = (int)((1l * all - ans) % mod);
  					//all = ((all % mod) + mod) % mod;
  					
  					pw.println(ans);
  				}
  			}
  			
  		}
  		
  		pw.flush();
  	}
  
  	public static class Scanner {
  		StringTokenizer st;
  		BufferedReader br;
  
  		public Scanner(InputStream system) {
  			br = new BufferedReader(new InputStreamReader(system));
  		}
  
  		public Scanner(String file) throws Exception {
  			br = new BufferedReader(new FileReader(file));
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public char nextChar() throws IOException {
  			return next().charAt(0);
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public int[] nextIntArray(int n) throws IOException {
  			int[] array = new int[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextInt();
  			return array;
  		}
  
  		public Integer[] nextIntegerArray(int n) throws IOException {
  			Integer[] array = new Integer[n];
  			for (int i = 0; i < n; i++)
  				array[i] = new Integer(nextInt());
  			return array;
  		}
  
  		public long[] nextLongArray(int n) throws IOException {
  			long[] array = new long[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextLong();
  			return array;
  		}
  
  		public double[] nextDoubleArray(int n) throws IOException {
  			double[] array = new double[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextDouble();
  			return array;
  		}
  
  		public static int[] shuffle(int[] a) {
  			int n = a.length;
  			Random rand = new Random();
  			for (int i = 0; i < n; i++) {
  				int tmpIdx = rand.nextInt(n);
  				int tmp = a[i];
  				a[i] = a[tmpIdx];
  				a[tmpIdx] = tmp;
  			}
  			return a;
  		}
  
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  
  		public void waitForInput() throws InterruptedException {
  			Thread.sleep(3000);
  		}
  	}
  }
  