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 static java.util.Arrays.*;
  
  import java.math.*;
  
  public class C {
  
  	public Object solve () {
  		int N = sc.nextInt();
  		int [] C = sc.nextInts();
  
  		int M = maximum(C);
  		int A = 0, B = 0;
  		for (int x : C)
  			if (x == M)
  				++A;
  			else if (x == M-1)
  				++B;
  
  		if (A > 1)
  			return F[N];
  
  		if (B == 0)
  			return 0;
  
  		long res = mod(F[N] * modInv(F[B+1]));
  		res = mod(res * (F[B+1] - F[B]));
  		return res;
  	}
  
  	private static final int CONTEST_TYPE = 2;
  
  	private static long [] F;
  
  	private static void init () {
  		F = modFact(200010);
  	}
  
  	private static final int MOD = 998244353;
  	/**
  	 * For every i in [0,N), bits[i] is the list of i's bits, in increasing order.
  	 */
  	int [][] bits (int N) {
  		int M = 1 << N;
  		int [][] res = new int [M][];
  		for (int m = 0; m < M; ++m) {
  			res[m] = new int [Integer.bitCount(m)];
  			int t = m;
  			for (int i = 0; i < res[m].length; ++i) {
  				int j = (t & (-t));
  				res[m][i] = Integer.numberOfTrailingZeros(j);
  				t -= j;
  			}
  		}
  		return res;
  	}
  	private static int maximum (int ... A) { return stream(A).max().getAsInt(); }
  	private static long mod (long x) { return mod(x, MOD); }
  	private static long mod (long x, long mod) { return ((x % mod) + mod) % mod; }
  	private static long [] modFact (int N) {
  		long [] res = new long [N+1]; res[0] = 1;
  		for (int i = 1; i <= N; ++i)
  			res[i] = mod(res[i-1] * i);
  		return res;
  	}
  	private static long modInv (long x) { return modInv(x, MOD); }
  	private static long modInv (long x, int mod) { return BigInteger.valueOf(x).modInverse(BigInteger.valueOf(mod)).longValue(); }
  	//////////////////////////////////////////////////////////////////////////////////// OFF
  	private static IOUtils.MyScanner sc = new IOUtils.MyScanner();
  	private static class IOUtils {
  		public static class MyScanner {
  			public String next () { newLine(); return line[index++]; }
  			public int nextInt () { return Integer.parseInt(next()); }
  			public String nextLine () { line = null; return readLine(); }
  			public String [] nextStrings () { return split(nextLine()); }
  			public int [] nextInts () {
  				String [] S = nextStrings(); int N = S.length;
  				int [] res = new int [N];
  				for (int i = 0; i < N; ++i)
  					res[i] = Integer.parseInt(S[i]);
  				return res;
  			}
  			//////////////////////////////////////////////
  			private boolean eol () { return index == line.length; }
  			private String readLine () {
  				try {
  					int b;
  					StringBuilder res = new StringBuilder();
  					while ((b = System.in.read()) < ' ')
  						continue;
  					res.append((char)b);
  					while ((b = System.in.read()) >= ' ')
  						res.append((char)b);
  					return res.toString();
  				} catch (Exception e) {
  					throw new Error (e);
  				}
  			}
  			private MyScanner () {
  				try {
  					while (System.in.available() == 0)
  						Thread.sleep(1);
  					start();
  				} catch (Exception e) {
  					throw new Error(e);
  				}
  			}
  			private String [] line;
  			private int index;
  			private void newLine () {
  				if (line == null || eol()) {
  					line = split(readLine());
  					index = 0;
  				}
  			}
  			private String [] split (String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
  		}
  		private static String build (Object o, Object ... A) { return buildDelim(" ", o, A); }
  		private static String buildDelim (String delim, Object o, Object ... A) {
  			java.util.List<String> str = new java.util.ArrayList<>();
  			append(str, o);
  			for (Object p : A)
  				append(str, p);
  			return String.join(delim, str.toArray(new String [0]));
  		}
  		//////////////////////////////////////////////////////////////////////////////////
  		private static java.text.DecimalFormat formatter = new java.text.DecimalFormat("#.#########");
  		private static void start () { if (t == 0) t = millis(); }
  		private static void append (java.util.function.Consumer<Object> f, java.util.function.Consumer<Object> g, Object o) {
  			if (o.getClass().isArray()) {
  				int len = java.lang.reflect.Array.getLength(o);
  				for (int i = 0; i < len; ++i)
  					f.accept(java.lang.reflect.Array.get(o, i));
  			}
  			else if (o instanceof Iterable<?>)
  				((Iterable<?>)o).forEach(f::accept);
  			else
  				g.accept(o instanceof Double ? formatter.format(o) : o);
  		}
  		private static void append (java.util.List<String> str, Object o) {
  			append(x -> append(str, x), x -> str.add(x.toString()), o);
  		}
  		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
  		private static void flush () { pw.flush(); System.out.flush(); }
  		private static void print (Object o, Object ... A) {
  			String res = build(o, A);
  			if (DEBUG == 2)
  				err(res, '(', time(), ')');
  			if (res.length() > 0)
  				pw.println(res);
  			if (DEBUG == 1)
  				flush();
  		}
  		private static void err (Object o, Object ... A) { System.err.println(build(o, A)); }
  		private static int DEBUG;
  		private static void exit () {
  			String end = "------------------" + System.lineSeparator() + time();
  			switch(DEBUG) {
  			case 1: print(end); break;
  			case 2: err(end); break;
  			}
  			IOUtils.pw.close();
  			System.out.flush();
  			System.exit(0);
  		}
  		private static long t;
  		private static long millis () { return System.currentTimeMillis(); }
  		private static String time () { return "Time: " + (millis() - t) / 1000.0; }
  		private static void run (int N) {
  			try { DEBUG = Integer.parseInt(System.getProperties().get("DEBUG").toString()); }
  			catch (Throwable t) {}
  
  			for (int n = 1; n <= N; ++n) {
  				Object res = new C().solve();
  				if (res != null) {
  					@SuppressWarnings("all")
  					Object o = CONTEST_TYPE == 0 ? "Case #" + n + ": " + build(res) : res;
  					print(o);
  				}
  			}
  			exit();
  		}
  	}
  	////////////////////////////////////////////////////////////////////////////////////
  	public static void main (String[] args) {
  		@SuppressWarnings("all")
  		int N = CONTEST_TYPE == 1 ? 1 : sc.nextInt();
  		init();
  		IOUtils.run(N);
  	}
  }
  