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.*;
  import java.math.*;
  import java.util.*;
  
  
  // @author : Dinosparton 
  
  public class test {
  	 
  	   static class Pair{ 
  		   long x;
  		   long y;
  		   
  		   Pair(long x,long y){ 
  			   this.x = x;
  			   this.y = y;
  			   
  		   }
  	   }
  	  
  	   static class Sort implements Comparator<Pair>
  	   {
  
  	       @Override
  	       public int compare(Pair a, Pair b)
  	       {
  	           if(a.x!=b.x)
  	           {
  	               return (int)(a.x - b.x);
  	           }
  	           else
  	           {
  	               return (int)(a.y-b.y);
  	           }
  	       }
  	   }
  	   
  	   static class Compare { 
  		   
  		     void compare(Pair arr[], int n) 
  		    { 
  		        // Comparator to sort the pair according to second element 
  		        Arrays.sort(arr, new Comparator<Pair>() { 
  		            @Override public int compare(Pair p1, Pair p2) 
  		            { 
  		            	if(p1.x!=p2.x) {
  		                return (int)(p1.x - p2.x); 
  		            	}
  		            	else { 
  		            		return (int)(p1.y - p2.y);
  		            	}
  		            } 
  		        }); 
  		  
  //		        for (int i = 0; i < n; i++) { 
  //		            System.out.print(arr[i].x + " " + arr[i].y + " "); 
  //		        } 
  //		        System.out.println(); 
  		    } 
  		} 
  	 
  	   static class Scanner {
  	        BufferedReader br;
  	        StringTokenizer st;
  	 
  	        public Scanner()
  	        {
  	            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 = 998244353;
  	   static long mul(long x, long y){
  	        return (x * y) % mod;
  	    }
  	   static long bin_pow(long x,long p){
  	        if(p == 0)  return 1;
  	        if((p & 1)!=0)   return mul(x, bin_pow(x, p - 1));
  	        return bin_pow(mul(x,x), p / 2);
  	    }
  	    static long rev(long  x) {
  	        return bin_pow(x, mod - 2);
  	    }
  	    static long div(long x, long y){
  	        return mul(x, rev(y));
  	    }
  	    static long add(long x,long y){
  	        x += y;
  	        if(x >= mod) return (x % mod);
  	        return x;
  	    }
  	    static long sub(long x, long y){
  	        x -= y;
  	        if(x < 0)   return (x + mod);
  	        return x;
  	    }
  	    
  	   public static void main(String args[]) throws Exception { 
  		
  		   
  		   Scanner sc = new Scanner();
  		   StringBuilder res = new StringBuilder();
  		
  		   long fac[]=new long[200005];
  	        fac[0]=fac[1]=1;
  	        for(int i=2;i<fac.length;i++){
  	            fac[i]=mul(i,fac[i-1]);
  	        }
  	        
  		   int tc = sc.nextInt();
  		   
  		   while(tc-->0) { 
  			
  			   int n = sc.nextInt();
  	            long max=Long.MIN_VALUE;
  	            
  	            HashMap<Long,Long> map=new HashMap<>();
  	            
  	            for(int i=0;i<n;i++){
  	            	
  	                long curr=sc.nextLong();
  	                
  	                if(map.containsKey(curr)){
  	                    map.put(curr,map.get(curr)+1);
  	                }else{
  	                    map.put(curr,(long)1);
  	                }
  	                
  	                max=Math.max(max,curr);
  	            }
  	 
  	            if(map.get(max)>1){
  	                res.append(fac[n]+"\n");
  	            }else if(map.get(max)==1){
  	                if(map.containsKey(max-1)==false){
  	                    res.append(0+"\n");
  	                }else{
  	                    long k=map.get(max-1);
  	                    long ans=sub(fac[n],div(fac[n],k+1));
  	                    res.append(ans+"\n");
  	                }
  	            }
  	 
  		   }
  			System.out.println(res);
  		   
  	   }
  }  
  
  
  
  
  