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: /*==========================================================================
  *   	AUTHOR:    RonWonWon
  *   	CREATED:   08.09.2021 23:41:43        
  /*==========================================================================*/
  import java.io.*;
  import java.util.*;
  
  public class C {
  	
  	public static void main(String[] args) throws IOException {
  		/*
      in.ini();
      PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
      */
      long startTime = System.nanoTime();
      int t = in.nextInt(), T = 1;  
      while(t-->0) { 
        HashMap<Integer, Integer> freq = new HashMap<>();
        int n = in.nextInt();
        int a[] = in.readArray(n);
        for(int i : a) freq.put(i, freq.getOrDefault(i,0)+1);
        int m1 = 0, m2 = -1;
        for(int i : freq.keySet()) m1 = Math.max(m1, i);
        for(int i : freq.keySet()) if(m1!=i) m2 = Math.max(m2, i);
        if(freq.get(m1)>1){
        	long ans = 1;
        	for(int i=1;i<=n;i++) ans = (ans*i)%mod;
        	out.println(ans);
        }
        else{
        	if(m2==-1) out.println(0);
        	else if(m1-m2!=1) out.println(0);
        	else{
        		long ans = 0;
        		f = new long[n+1];
        		f[0] = 1; f[1] = 1;
        		for(int i=2;i<=n;i++) f[i] = (f[i-1] * i)%mod;
        		for(int i=0;i<=n-2;i++){
        			if(i<freq.get(m2)) ans = (ans + f[n-1])%mod;
        			else{
        				int diff = (i-freq.get(m2));
        				long ncr = nCk(n-freq.get(m2)-1,diff);
        				long minus = (f[i]*ncr)%mod;
        				minus = (minus*f[n-i-1])%mod;
        				ans = (ans + (f[n-1] - minus + mod)%mod)%mod;
        			}
        		}
        		out.println(ans);
        	}
        }  
     		//out.println("Case #"+T+": "+ans); T++;
      }
      /*
      startTime = System.nanoTime() - startTime;
      System.err.println("Time: "+(startTime/1000000));
      */
      out.flush();
  	}
  	
  	static final long mod=998244353;
  	static long f[];
  	
  	static long mul(long a, long b) {
  		return a*b%mod;
  	}
  
  	static long fastPow(long base, long exp) {
  		if (exp==0) return 1;
  		long half=fastPow(base, exp/2);
  		if (exp%2==0) return mul(half, half);
  		return mul(half, mul(half, base));
  	}
  	
  	static long modInv(long x) {
  		return fastPow(x, mod-2);
  	}
  	
  	static long fact(int x) {
  		return f[x];
  	}
  	
  	static long nCk(int n, int k) {
  		return mul(fact(n), mul(modInv(fact(k)), modInv(fact(n-k))));
  	}
  	
  	static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(System.out);
  	static int oo = Integer.MAX_VALUE;
  	static long ooo = Long.MAX_VALUE;
  	
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		
  		void ini() throws FileNotFoundException {
  			br = new BufferedReader(new FileReader("input.txt"));
  		}
  		
  		String next() {
  			while(!st.hasMoreTokens())
  				try { st = new StringTokenizer(br.readLine()); }
  				catch(IOException e) {}
  			return st.nextToken();
  		}
  		
  		String nextLine(){
  			try{ return br.readLine(); } 
  			catch(IOException e) { } return "";
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		int[] readArray(int n) {
  			int a[] = new int[n];
  			for(int i=0;i<n;i++) a[i] = nextInt();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		
  		long[] readArrayL(int n) {
  			long a[] = new long[n];
  			for(int i=0;i<n;i++) a[i] = nextLong();
  			return a;
  		}
  		
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		
  		double[] readArrayD(int n) {
  			double a[] = new double[n];
  			for(int i=0;i<n;i++) a[i] = nextDouble();
  			return a;
  		}
  	}
  
  	static final Random random = new Random();
  
  	static void ruffleSort(int[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n), temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  	
  	static void ruffleSortL(long[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n);
  			long temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  }
  