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: // package codeforce.Training1900;
  
  import java.io.PrintWriter;
  import java.util.*;
  
  public class JuryMeeting {
      //    MUST SEE BEFORE SUBMISSION
  //    check whether int part would overflow or not, especially when it is a * b!!!!
  //    check if top down dp would cause overflow or not !!!!!!!!!!!!!!!!!!!!!!
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int t = 1;
          t = sc.nextInt();
          for (int i = 0; i < t; i++) {
              solve(sc, pw);
          }
          pw.close();
      }
      static int mod = 998244353;
      static Combinatorial cb = new Combinatorial(mod, (int) 2e5 + 5);
      static void solve(Scanner in, PrintWriter out){
          int n = in.nextInt();
          int[] a = new int[n];
          for (int i = 0; i < n; i++) {
              a[i] = in.nextInt();
          }
          Map<Integer, Integer> mp = new HashMap<>();
          int max = 0;
          for(int x : a) {
              mp.put(x, mp.getOrDefault(x, 0) + 1);
              max = Math.max(max, x);
          }
          if (mp.get(max) == 1 && mp.getOrDefault(max - 1, 0) == 0){
              out.println(0);
          }else{
              long tot = 1;
              for(int i = n; i >= 1; i--){
                  tot = (tot * i) % mod;
              }
              long[] pre = new long[n + 1];
              pre[1] = 1;
              for(int i = 2; i <= n; i++){
                  pre[i] = (pre[i - 1] * i) % mod;
              }
              if (mp.get(max) >= 2){
                  out.println(tot);
              }else{
                  tot = 1;
                  for(int i = n - 1; i >= 1; i--){
                      tot = (tot * i) % mod;
                  }
                  tot = (tot * (n - 1)) % mod;
                  int rest = n - mp.get(max) - mp.get(max - 1);
                  long ad = 0;
                  for(int i = 1; i <= rest; i++){
                      long right = mut(cb.ncr(rest, i), pre[i]);
                      long left = pre[n - 1 - i];
                      ad = add(ad, mut(left, right));
                  }
                  tot -= ad;
                  if (tot < 0) tot += mod;
                  out.println(tot);
              }
          }
      }
      static long add(long a, long b){
          a += b;
          if (a >= mod) a-= mod;
          return a;
      }
      static long mut(long a, long b){
          return (a * b) % mod;
      }
      //      credits to SecondThread
      // Use this instead of Arrays.sort() on an array of ints. Arrays.sort() is n^2
      // worst case since it uses a version of quicksort. Although this would never
      // actually show up in the real world, in codeforces, people can hack, so
      // this is needed.
      static void ruffleSort(int[] a) {
          //ruffle
          int n=a.length;
          Random r=new Random();
          for (int i=0; i<a.length; i++) {
              int oi=r.nextInt(n), temp=a[i];
              a[i]=a[oi];
              a[oi]=temp;
          }
  
          //then sort
          Arrays.sort(a);
      }
      // credits to cuiaoxiang
      static class Combinatorial {
  
          int MOD;
          int n;
          long[] fact;
          long[] inv;
          long[] ifact;
          public Combinatorial(int mod, int n) {
              this.MOD = mod;
              this.n = n;
              fact = new long[n + 1];
              inv = new long[n + 1];
              ifact = new long[n + 1];
              inv[1] = 1;
              for (int i = 2; i < n; ++i) {
                  inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
              }
              fact[0] = ifact[0] = 1;
              for (int i = 1; i < n; ++i) {
                  fact[i] = fact[i - 1] * i % MOD;
                  ifact[i] = ifact[i - 1] * inv[i] % MOD;
              }
  
          }
  
          long ncr(int n, int m){
              if (n < m || m < 0) return 0;
              return fact[n] * ifact[m] % MOD * ifact[n - m] % MOD;
          }
      }
  
  }
  