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.InputStreamReader;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  
  
  public class JuryMeeting {
  
      private static final long MOD=998244353;
      private static final int mx = 2 * 100001;
      private static final long[] fact = new long[mx];
      private static final long[] inv = new long[mx];
      private static final long[] invfact = new long[mx];
      private static InputStreamReader ina = new InputStreamReader(System.in);
      private static BufferedReader in = new BufferedReader(ina);
  
      public static void main(String[] args )throws Exception {
  
          fillFactorials();
          int teste = Integer.parseInt(in.readLine());
          while (teste-- > 0) {
              solve();
          }
      }
  
      private static void fillFactorials() {
          fact[0] = fact[1] = inv[1] = invfact[0] = invfact[1] = 1;
          for (int i = 2; i < mx; i++) {
              fact[i] = (fact[i-1]*i) % MOD;
              inv[i] = MOD - MOD/i*inv[(int)MOD%i]%MOD;
              invfact[i] = inv[i]*invfact[i-1]%MOD;
          }
      }
  
      private static void solve() throws Exception {
          int inimesi = Integer.parseInt(in.readLine());
          String[] tasks_str = in.readLine().split(" ");
          ArrayList<Long> tasks = new ArrayList<>();
          for (int i = 0; i < inimesi; i++) {
              tasks.add(Long.valueOf(tasks_str[i]));
          }
  
          long max_tasks = Collections.max(tasks);
          int max_taskide_arv = Collections.frequency(tasks, max_tasks);
          int max_miinus_1_taskide_arv = Collections.frequency(tasks, max_tasks-1);
          int ylejaanud = tasks.size() - max_taskide_arv - max_miinus_1_taskide_arv;
  
          // Kui max_taskide arvuga inimesi on rohkem kui 1 siis on kõik permutatsiooni ilusad, sest nad omavahel saavad hakkama
          if(max_taskide_arv > 1){
              System.out.println(fact[tasks.size()]);
          } else if (max_miinus_1_taskide_arv == 0){ // Kui max_taskidega on 1 inimene, siis peab leiduma vähemnalt 1 inimene kellel on taske max_taskide_arv - 1
              System.out.println("0");
          } else {
              // Kõik võimalikud
              long permutatsioonid = fact[tasks.size()];
  
              // Nüüd hakkame maha lahutama neid mis ei sobi e kus kõik max_miinus_1_taskide_arv on eespool
              // Niipalju kui on ylejäänuuid tuleb kombineerida
              // 3 3 3 4 1 2 kõik max miinus-1 ees
              // 3 3 3 1 4 2 kõik max miinus-1 1 ees ja 1 ylejäänud ka
              // 3 3 3 1 2 4
              for (int i = 0; i <= ylejaanud; i++) {
  
                  // Kõik miinus-1 on eesotsas ja lisaks ylejäänutest i
                  long eesmise_otsa_perm = fact[max_miinus_1_taskide_arv + i];
  
                  // Taga permuteeruvad ylejäänud - u
                  long tagumise_otsa_perm = fact[ylejaanud - i];
  
                  // Lisaks kombinatsioonid mida saab teha ylejäänutes koguarvust niipalju kaupa kui on ees otas
                  // eesotsas on i tükki, järelikult i kaupa ylejaannu-st
                  // https://cp-algorithms.com/combinatorics/binomial-coefficients.html
  
                  long kombinatsioonid = kombinatsioonid(ylejaanud, i);
  
                  // Pane tähele, et vahepeal tuleb %MOD teha, muidu "jookseb üle"
                  permutatsioonid = (permutatsioonid - ((eesmise_otsa_perm * tagumise_otsa_perm)%MOD * kombinatsioonid)%MOD + MOD) % MOD;
              }
              System.out.println(permutatsioonid);
          }
          // precomp(25);
      }
  
      private static long kombinatsioonid(int millest, int kaupa) {
          return fact[millest] * invfact[millest-kaupa] % MOD * invfact[kaupa] % MOD;
      }
  
  
      private static long nCrModp(int n, int r)
      {
          if (r > n - r)
              r = n - r;
  
          long[] C = new long[r+1];
  
          C[0] = 1;
  
          for (int i = 1; i <= n; i++) {
              for (int j = Math.min(i, r); j > 0; j--)
                  C[j] = (C[j] + C[j - 1]) % MOD;
          }
          return C[r];
      }
  
  
  }
  