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.BufferedWriter;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.OutputStreamWriter;
  import java.util.*;
  public class c731 {
  
  	public static void main(String[] args) throws IOException {
  //try {
  FastReader sc = new FastReader();
  BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
  int t = sc.nextInt();
  int m = 998244353;
  long[] fac = new long[200001];
  fac[0] =1;
  fac[1] = 1;
  for(int i = 2 ; i<200001;i++) {
  	fac[i] = (fac[i-1]*i)%m;
  }
  //for(int i = 0 ; i<22; i++) {
  //	System.out.print(fac[i] + " ");
  //}
  for(int o = 0 ; o<t;o++) {
  int n = sc.nextInt();
  long[] arr = new long[n];
  for(int i =  0 ; i<n;i++) {
  	arr[i] = sc.nextLong();
  }
  shuffleArray(arr);
  Arrays.sort(arr);
  if(arr[n-1]==arr[n-2]) {
  	System.out.println(fac[n]);
  }else if(arr[n-1]-arr[n-2]>1) {
  	System.out.println(0);
  }else {
  	int x = 0;
  	int j = n-2;
  	while(j>=0 && arr[n-1] - arr[j] == 1) {
  		x ++;
  	j--;
  	}
  	int y = n-x-1;
  	long val = 0;
  //	System.out.println("y = " + y);
  	for(int i = 1 ; i<=y;i++) {
  //		val = (val%m + (((ncr(y, i, m, fac)*fac[i])%m)%m*fac[n-i-1]%m)%m + m)%m;
  		long c =  mul(mul(ncr(y,i,m,fac),fac[i],m),fac[n-i-1],m);
  	val = add(val,c,m);
  //	System.out.println("ncr = " + ncr(y,i,m,fac));
  //	System.out.println("val = " + val);
  	}
  	long ans = (fac[n]%m - fac[n-1]%m + m)%m;
  	ans = (ans%m - val%m + m) % m;
  	System.out.println(ans);
  //	System.out.println((fac[n]%m-(ncr(n,x+1,m,fac)%m*fac[x]%m*fac[n-x-1]%m + m)%m));
  //long h = mul((mul(ncr(n,x+1,m,fac),fac[x],m)),fac[n-x-1],m);
  //System.out.println((fac[n]%m - h %m + m)%m);
  
  }
  
  
  }
  
  	}
  	
  	public static long add(long a , long b , int m) {
  		return(a%m + b%m + m)%m;
  	}
  	public static long mul(long a , long b , int m) {
  		return (a* b) %m;
  	}
  	
  	public static long power(long x, int y, int p)
  	    {
  	 
  	        long res = 1;
  	 
  	        x = x % p;
  	 
  	        while (y > 0) {
  	 
  	            if (y % 2 == 1)
  	                res = (res * x) % p;
  	 
  	            y = y >> 1; 
  	            x = (x * x) % p;
  	        }
  	 
  	        return res;
  	    }
  	 
  	    public static long modInverse(long n, int p)
  	    {
  	        return power(n, p - 2, p);
  	    }
  	 
  	   public static long ncr(int n, int r,
  	                             int p,long[] fac)
  	    {
  	 
  	          if (n<r)
  	              return 0;
  	        if (r == 0)
  	            return 1;
  	 
  	   
  //	        int[] fac = new int[n + 1];
  //	        fac[0] = 1;
  //	 
  //	        for (int i = 1; i <= n; i++)
  //	            fac[i] = fac[i - 1] * i % p;
  	 
  	        return (fac[n] * modInverse(fac[r], p)
  	                % p * modInverse(fac[n - r], p)
  	                % p)
  	            % p;
  	    }
  	 public static  void shuffleArray(long[] arr){
  	        int n = arr.length;
  	        Random rnd = new Random();
  	        for(int i=0; i<n; ++i){
  	            long tmp = arr[i];
  	            int randomPos = i + rnd.nextInt(n-i);
  	            arr[i] = arr[randomPos];
  	            arr[randomPos] = tmp;
  	        }   
  	}
  
  }
  
  
  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;
      }
  }