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 Jury_Meeting {
  
  	static 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();
  		}
  
  		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;
  		}
  
  	}
  
  	public static void shuffle(long[] a) {
  		Random r = new Random();
  
  		for (int i = 0; i <= a.length - 2; i++) {
  			int j = i + r.nextInt(a.length - i);
  
  			swap(a, i, j);
  		}
  
  		Arrays.sort(a);
  	}
  
  	public static void swap(long[] a, int i, int j) {
  		long temp = a[i];
  		a[i] = a[j];
  		a[j] = temp;
  	}
  
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		FastReader t = new FastReader();
  		PrintWriter o = new PrintWriter(System.out);
  		int test = t.nextInt();
  		long mod = 998244353;
  		long[] fac = new long[200001];
  		fac[0] = 1;
  
  		for (int i = 1; i < fac.length; ++i) {
  			fac[i] = (fac[i - 1] * i) % mod;
  		}
  
  		while (test-- > 0) {
  			int n = t.nextInt();
  			long[] a = new long[n];
  
  			for (int i = 0; i < n; ++i)
  				a[i] = t.nextLong();
  
  			shuffle(a);
  
  			if (a[n - 1] == a[n - 2])
  				o.println(fac[n]);
  			else if (a[n - 1] >= a[n - 2] + 2)
  				o.println("0");
  			else {
  				long k = 0;
  
  				for (long v : a)
  					if (a[n - 2] == v)
  						++k;
  
  				long nCr = fun(n, k + 1, mod, fac);
  				long invalid = ((nCr * fac[(int) k]) % mod) * fac[(int) (n - k - 1)] % mod;
  				long ans = (fac[n] - invalid + mod) % mod;
  
  				o.println(ans);
  			}
  		}
  
  		o.flush();
  		o.close();
  	}
  
  	private static long power(long x, long y, long p) {
  		long res = 1;
  		x = x % p;
  
  		while (y > 0) {
  			if (y % 2 == 1)
  				res = (res * x) % p;
  
  			y = y >> 1;
  			x = (x * x) % p;
  		}
  
  		return res;
  	}
  
  	private static long modInverse(long n, long p) {
  		return power(n, p - 2, p);
  	}
  
  	private static long fun(long n, long r, long mod, long[] fac) {
  		if (n < r)
  			return 0;
  
  		if (r == 0)
  			return 1;
  
  		return (fac[(int) n] * modInverse(fac[(int) r], mod) % mod * modInverse(fac[(int) (n - r)], mod) % mod) % mod;
  	}
  }