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.io.PrintWriter;
  import java.util.HashMap;
  import java.util.StringTokenizer;
  
  public class meeting {
  	long mod = 998244353;
  
  	public static void main(String[] args) {
  		FastScanner in = new FastScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int tt = in.nextInt();
  		for (int tc = 0; tc < tt; tc++) {
  			meeting solver = new meeting();
  			solver.solve(in, out);
  		}
  		out.flush();
  	}
  
  	void solve(FastScanner in, PrintWriter out) {
  		int n = in.nextInt();
  		long a[] = new long[n];
  		long mx = 0;
  		HashMap<Long, Integer> m = new HashMap<>();
  		for (int i = 0; i < n; i++) {
  			a[i] = in.nextLong();
  			mx = Math.max(a[i], mx);
  			m.put(a[i], m.getOrDefault(a[i], 0) + 1);
  		}
  		long f[] = new long[n + 1];
  		f[0] = 1;
  		for (int i = 1; i <= n; i++) f[i] = mul(i, f[i - 1]);
  		if (m.get(mx) > 1) {
  			out.println(f[n]);
  		} else {
  			if (m.containsKey(mx - 1)) {
  				int cnt2 = m.get(mx - 1);
  				long C2[] = new long[n + 1];
  				C2[cnt2] = f[cnt2];
  				long ans = C2[cnt2];
  				for (int i = cnt2 + 1; i < n; i++) {
  					C2[i] = mul(div(i, sub(i, cnt2)), C2[i - 1]);
  					ans = add(ans, C2[i]);
  				}
  				ans = mul(ans, f[n - 1 - cnt2]);
  				ans = sub(f[n], ans);
  				out.println(ans);
  			} else {
  				out.println(0);
  			}
  		}
  	}
  
  	long mul(long x, long y) { return ((x * y) % mod + mod) % mod; }
  	long add(long x, long y) { return ((x + y) % mod + mod) % mod; }
  	long sub(long x, long y) { return ((x - y) % mod + mod) % mod; }
  	long exp(long x, long y) {
  		long res = 1;
  		while (y > 0) {
  			if (y % 2 == 1) res = mul(res, x);
  			x = mul(x, x);
  			y /= 2;
  		}
  		return res;
  	}
  	long div(long x, long y) { return mul(x, exp(y, mod - 2)); }
  
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer tok = new StringTokenizer("");
  		String next() {
  			while(!tok.hasMoreTokens()) {
  				try {
  					tok = new StringTokenizer(br.readLine());
  				} catch (Exception e) {
  					e.printStackTrace();
  				}
  			}
  			return tok.nextToken();
  		}
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		float nextFloat() {
  			return Float.parseFloat(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  }
  