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.File;
  import java.io.FileNotFoundException;
  import java.io.FileReader;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.InputStreamReader;
  import java.io.OutputStream;
  import java.io.PrintWriter;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.LinkedList;
  import java.util.List;
  import java.util.Queue;
  import java.util.StringTokenizer;
  import java.util.TreeMap;
  
  
  public class Main {
  
  	static class Task {
  		
  		int NN = 200005;
  		int MOD = 998244353;
  		int INF = 2000000000;
  		long INFINITY = 2000000000000000000L;
  
  		public void solve(InputReader in, PrintWriter out) {
  			fac[0] = 1;
  			for(int i=1;i<200005;++i) fac[i] = (fac[i - 1] * 1L * i)%MOD;
  			int t = in.nextInt();
  			while(t-->0){
  				int n = in.nextInt();
  				List<Integer> L = new ArrayList<>();
  				for(int i=0;i<n;++i){
  					L.add(in.nextInt());
  				}
  				Collections.sort(L);
  				int d = L.get(n - 1) - L.get(n - 2);
  				int c = 0;
  				for(int i=n-2;i>=0 && L.get(i).equals(L.get(n - 2));--i)++c;
  				if(d == 0) {
  					out.println(fac[n]);
  				} else if(d == 1) {
  					out.println((fac[n] - (fac[n] * exp(c+1, MOD-2))%MOD + MOD)%MOD);
  				} else {
  					out.println(0);
  				}
  			}
  		}
  
  		long exp(long x, long n) {
  			long ret = 1;
  			long sq = x;
  			while(n>0){
  				if(n%2!=0){
  					ret = (ret * sq) % MOD;
  				}
  				n /= 2;
  				sq = (sq*sq)%MOD;
  			}
  			return ret;
  		}
  
  		long [] fac = new long[200005];
  
  	}
  	
  	static void prepareIO(boolean isFileIO) {
  		//long t1 = System.currentTimeMillis();
  		Task solver = new Task();
  		// Standard IO
  		if(!isFileIO) { 
  			InputStream inputStream = System.in;
  	        OutputStream outputStream = System.out;
  	        InputReader in = new InputReader(inputStream);
  	        PrintWriter out = new PrintWriter(outputStream);
  	        solver.solve(in, out);
  	        //out.println("time(s): " + (1.0*(System.currentTimeMillis()-t1))/1000.0);
  	        out.close();
  		}
          // File IO
  		else {
  			String IPfilePath = System.getProperty("user.home") + "/Downloads/ip.in";
  	        String OPfilePath = System.getProperty("user.home") + "/Downloads/op.out";
  	        InputReader fin = new InputReader(IPfilePath);
  	        PrintWriter fout = null;
  	        try {
  				fout = new PrintWriter(new File(OPfilePath));
  			} catch (FileNotFoundException e) {
  				e.printStackTrace();
  			}
  	        solver.solve(fin, fout);
  	        //fout.println("time(s): " + (1.0*(System.currentTimeMillis()-t1))/1000.0);
  	        fout.close();
  		}
  	}
  	
  	public static void main(String[] args) {
          prepareIO(false);
  	}
  	
  	static class InputReader {
          public BufferedReader reader;
          public StringTokenizer tokenizer;
  
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
          
          public InputReader(String filePath) {
          	File file = new File(filePath);
              try {
  				reader = new BufferedReader(new FileReader(file));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
              tokenizer = null;
          }
          
          public String nextLine() {
          	String str = "";
          	try {
  				str = reader.readLine();
  			} catch (IOException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
          	return str;
          }
          
          public String next() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public long nextLong() {
              return Long.parseLong(next());
          }
  
          public double nextDouble() {
          	return Double.parseDouble(next());
          }
          
      }
  
  }