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.util.*;
  // import java.lang.*;
  import java.io.*;
  
  //           THIS TEMPLATE MADE BY AKSH BANSAL.
  
  public class Solution {
      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;
          }
      }
      private static boolean[] isPrime;
      private static void primes(){
          int num = (int)1e6; // PRIMES FROM 1 TO NUM
          isPrime = new boolean[num];
       
          for (int i = 2; i< isPrime.length; i++) {
             isPrime[i] = true;
          }
          for (int i = 2; i< Math.sqrt(num); i++) {
             if(isPrime[i] == true) {
                for(int j = (i*i); j<num; j = j+i) {
                   isPrime[j] = false;
                }
             }
          }
      }
      static void sort(int a[]){ // int -> long
          ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long
          for(int i=0;i<a.length;i++)
          arr.add(a[i]);
          Collections.sort(arr);
          for(int i=0;i<a.length;i++)
          a[i]=arr.get(i);
          
      }
      private static long gcd(long a, long b){
          if(b==0)return a;
          return gcd(b,a%b);
      }
      private static long pow(long x,long y){
          if(y==0)return 1;
          long temp = pow(x, y/2);
          if(y%2==1){
              return x*temp*temp;
          }
          else{
              return temp*temp;
          }
      }
  
      static PrintWriter out;
      static FastReader sc ;
      public static void main(String[] args) throws IOException {
          sc = new FastReader();
          out = new PrintWriter(System.out);
          // primes();
          // ________________________________
  
          int test = sc.nextInt();
          StringBuilder output = new StringBuilder();
  
          while (test-- > 0) {
              int n = sc.nextInt();
              int[] arr =new int[n];
              for(int i=0;i<n;i++){
                  arr[i ] =sc.nextInt();
              }
              output.append(solver(arr,n)).append("\n");
          }
  
          out.println(output);
          // _______________________________
  
          // int n = sc.nextInt();
          // out.println(solver());
          // ________________________________
          out.flush();
      }
      static int mod =  998244353;
      public static long solver(int[] arr, int n) {
          sort(arr);
          if(arr[n-1]-arr[n-2]>1){
              return 0;
          }
          int count = 1;
          int x = 1;
          for(int i=n-2;i>=0;i--){
              if(arr[n-1]==arr[i]){
                  count++;
                  x++;
              }
              else if(arr[n-1]==arr[i]+1){
                  x++;
              }
          }
          long res =1;
          // System.out.println("__"+ count);
          for(int i=1;i<=n;i++){
              if(count==1 && i==x){
                  res = (res*(i-1))%mod;
                  continue;
              }
              res = (res*i)%mod;
          }
          return res;
      }
  }