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: /*Everything is Hard 
   * Before Easy 
   * Jai Mata Dii 
   */ 
   
  import java.util.*;
  import java.io.*; 
    
  public class Main {
  	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; }} 
  //	static long mod = (long)(1e9+7); 
  	 static long mod = 998244353; 
  //	 static Scanner sc = new Scanner(System.in); 
  	static FastReader sc = new FastReader(); 
  	static PrintWriter out = new PrintWriter(System.out);
  	public static void main (String[] args) { 
  		int ttt = 1;
  		ttt = sc.nextInt();
  		precompute();
  		z :for(int tc=1;tc<=ttt;tc++){
  			int n = sc.nextInt();
  			int a[] = new int[n];
  			for(int i=0;i<n;i++) {
  				a[i] = sc.nextInt();
  			}
  			sort(a);
  			int max = 0, cnt = 0, smax = 0;
  			for(int i=0;i<n;i++) {
  				if(a[i]>max) {
  					smax = max;
  					max = a[i];
  					cnt = 1;
  				}
  				else if(a[i] == max) {
  					cnt++;
  				}
  				else if(smax > a[i]) {
  					smax = a[i];
  				}
  			}
  			if(cnt == 1 && (max - smax)>1) {
  				out.write("0\n");
  				continue;
  			}
  			long ans = 1;
  			for(long i = 1;i<=n;i++) {
  				ans = (ans * i)%mod;
  			}
  			if(cnt == 1) {
  				int less = 0;
  				for(int i=0;i<n;i++) {
  					if(a[i]<max-1) less++;
  				}
  				for(int i=0;i<=less;i++) {
  					long mul = ((fac[(int)less] * power(fac[(int)(less-i)],mod-2,mod))%mod * power(fac[(int)i],mod-2,mod)%mod)%mod;
  //					System.out.println(i+" "+mul);
  					long toMinus = 1L*mul*(fac[i]*fac[n-i-1]%mod)%mod;
  					ans = (ans - toMinus + mod)%mod;
  				}
  			}
  			out.write(ans+"\n");
  		}
  		out.close();
  	}
  	static long fac[] = new long[300006];
  	
  	static void precompute() {
  		int n = 300005;
  		fac[0] = 1;
  		for(int i=1;i<=n;i++)
  			fac[i] = (i * fac[i-1])%mod;
  	}
  	
  	static long power(long base, long p, long mod2) {
  		
  		long ans = 1;
  		while(p!=0) {
  			if(p%2==0) {
  				base = (base*base)%mod2;
  				p/=2;
  			}
  			else {
  				ans = (ans*base)%mod2;
  				p-=1;
  			}
  		}
  		return (ans%mod2);
  	}
  	static long pow(long a, long b){long ret = 1;while(b>0){if(b%2 == 0){a = (a*a)%mod;b /= 2;}else{ret = (ret*a)%mod;b--;}}return ret%mod;}
  	static long gcd(long a,long b){if(b==0) return  a; return gcd(b,a%b); } 
  	private static void sort(int[] a) {List<Integer> k = new ArrayList<>();for(int val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  	private static void ini(List<Integer>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void init(List<int[]>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void sort(long[] a) {List<Long> k = new ArrayList<>();for(long val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  }