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.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  import java.io.BufferedReader;
  import java.util.*;
  
  public class C {
  	private static final FastReader fs = new FastReader();
  	private static final PrintWriter out = new PrintWriter(System.out);
  	private static long m = 1_000_000_009;
  	private static long p = 31;
  	private static long mod = 998244353;
  	private static long[] perm;
  	
  	private static void fill() {
  		perm = new long[2_000_10];
  		perm[0] = 1;
  		for(int i = 1; i < 2_000_10; i++)
  			perm[i] = (i * perm[i-1]) % mod;
  	}
  	
  	private static void solve() {
  		int n = fs.nextInt();
  		int[]a = fs.readintarray(n);
  		int max = Arrays.stream(a).max().getAsInt();
  		long ans = perm[n];
  		if(Arrays.stream(a).filter(ele -> ele > max-2).count() > 1) {
  			if(Arrays.stream(a).filter(ele->ele == max).count() > 1)
  				out.println(ans);
  			else {
  				long count = Arrays.stream(a).filter(e -> e <= max-2).count();
  				ans = sub(ans, perm[n-1]);
  				for(int i = 1; i <= count; i++) {
  					ans = sub(ans, mul(mul(perm[(int) count], mPower(perm[(int) (count-i)], mod-2)), perm[n-i-1]));
  				}
  				out.println(ans);
  			}
  		}
  		else
  			out.println(0);
  	}
   
  	public static void main(String[] args) {
  		Thread t = new Thread(null, null, "", 1 << 28) {
  			public void run() {
  				fill();
  				int test_case = 1;
  				test_case = fs.nextInt();
  				for (int cs = 1; cs <= test_case; cs++)
  					solve();
  				out.close();
  			}
  		};
  		t.start();
  		try {
  			t.join();
  		} catch (InterruptedException e) {
  			e.printStackTrace();
  		}
  	}
   
  	private 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;
  		}
   
  		int[] readintarray(int n) {
  			int res[] = new int[n];
  			for (int i = 0; i < n; i++)
  				res[i] = nextInt();
  			return res;
  		}
   
  		long[] readlongarray(int n) {
  			long res[] = new long[n];
  			for (int i = 0; i < n; i++)
  				res[i] = nextLong();
  			return res;
  		}
  	}
  	
  	private static int max(int... arg) {
  		int length = arg.length;
  		if(length == 0)
  			throw null;
  		int maxVal = arg[0];
  		for(int i = 1; i < length; i++)
  			maxVal = Math.max(maxVal, arg[i]);
  		return maxVal;
  	}
  	
  	private static int min(int... arg) {
  		int length = arg.length;
  		if(length == 0)
  			throw null;
  		int maxVal = arg[0];
  		for(int i = 1; i < length; i++)
  			maxVal = Math.min(maxVal, arg[i]);
  		return maxVal;
  	}
  	
  	private static long max(long... arg) {
  		int length = arg.length;
  		if(length == 0)
  			throw null;
  		long maxVal = arg[0];
  		for(int i = 1; i < length; i++)
  			maxVal = Math.max(maxVal, arg[i]);
  		return maxVal;
  	}
  	
  	private static long min(long... arg) {
  		int length = arg.length;
  		if(length == 0)
  			throw null;
  		long maxVal = arg[0];
  		for(int i = 1; i < length; i++)
  			maxVal = Math.min(maxVal, arg[i]);
  		return maxVal;
  	}
  	
  	private static int abs(int a) {
  		return a >= 0 ? a : -a;
  	}
  	
  	private static int abs(int a, int b) {
  		return a-b >= 0 ? a-b : b-a;
  	}
  	
  	private static long abs(long a) {
  		return a >= 0 ? a : -a;
  	}
  	
  	private static long abs(long a, long b) {
  		return a-b >= 0 ? a-b : b-a;
  	}
  	
  	private static int power(int base, int exponent) {
  		int res = 1;
  		while(exponent > 0) {
  			if((exponent & 1) != 0)
  				res *= base;
  			exponent >>= 1;
  			base *= base;
  		}
  		return res;
  	}
  	
  	private static long power(long base, long exponent) {
  		long res = 1;
  		while(exponent > 0) {
  			if((exponent & 1) != 0)
  				res *= base;
  			exponent >>= 1;
  			base *= base;
  		}
  		return res;
  	}
  	
  	private static long mPower(long base, long exponent) {
  		long res = 1;
  		base %= mod;
  		while(exponent > 0) {
  			if((exponent & 1) != 0)
  				res = mul(res, base);
  			exponent >>= 1;
  			base = mul(base, base);
  		}
  		return res;
  	}
  	
  	private static long mul(long a, long b) {
  		return a*b % mod;
  	}
  	
  	private static long add(long a, long b) {
  		return (a+b) % mod;
  	}
  	
  	private static long sub(long a, long b) {
  		long res = (a - b) % mod;
  		if(res < 0)
  			res += mod;
  		return res;
  	}
  }