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.util.ArrayList;
  import java.util.Arrays;
  import java.util.Random;
  import java.util.StringTokenizer;
  
  public class A {
  
      public static void main(String[] args) {
          FastReader scan = new FastReader();
          PrintWriter out = new PrintWriter(System.out);
          Task solver = new Task();
          int t = scan.nextInt();
          for(int tt = 1; tt <= t; tt++) solver.solve(tt, scan, out);
          out.close();
      }
  
      static class Task {
          static long[] fact, invFact, naturalInverse;
          static final long MOD = 998244353;
          static final int MAX = (int) (2e5 + 100);
  
          public void solve(int testNumber, FastReader scan, PrintWriter out) {
              if(testNumber == 1) precomp();
  
              int n = scan.nextInt();
              int[] a = new int[n];
  
              for(int i = 0; i < n; i++) a[i] = scan.nextInt();
              ruffleSort(a);
              if(a[n - 1] == a[n - 2]) out.println(fact[n]);
              else if(a[n - 1] > a[n - 2] + 1) out.println(0);
              else {
                  long ans = 0;
                  int count = 0;
  
                  for(int i = 0; i < n; i++) {
                      if(a[i] == a[n - 2]) count++;
                  }
                  for(int i = 0; i < n; i++) {
                      long total = fact[n - 1];
                      long bad = perm(i, count) * fact[n - 1 - count] % MOD;
                      total = ((total - bad) % MOD + MOD) % MOD;
                      ans = (ans + total) % MOD;
                  }
                  out.println(ans);
              }
          }
  
          void precomp() {
              fact = new long[MAX + 1];
              invFact = new long[MAX + 1];
              naturalInverse = new long[MAX + 1];
  
              fact[0] = invFact[0] = invFact[1] = naturalInverse[0] = naturalInverse[1] = 1;
              for(int i = 1; i <= MAX; i++) {
                  fact[i] = (fact[i-1]*i)%MOD;
                  if(i == 1) continue;
                  naturalInverse[i] = naturalInverse[(int) (MOD % i)] * (MOD - MOD/i) % MOD;
                  invFact[i] = (invFact[i-1]*naturalInverse[i])%MOD;
              }
          }
  
          static long perm(int a, int b) {
              if(a < b) return 0;
              return(fact[a]%MOD*invFact[a-b])%MOD;
          }
  
      }
  
      static void ruffleSort(int[] a) {
          Random get = new Random();
          for (int i = 0; i < a.length; i++) {
              int r = get.nextInt(a.length);
              int temp = a[i];
              a[i] = a[r];
              a[r] = temp;
          }
          Arrays.sort(a);
      }
  
      static void ruffleSort(long[] a) {
          Random get = new Random();
          for (int i = 0; i < a.length; i++) {
              int r = get.nextInt(a.length);
              long temp = a[i];
              a[i] = a[r];
              a[r] = temp;
          }
          Arrays.sort(a);
      }
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          public FastReader(String s) throws FileNotFoundException {
              br = new BufferedReader(new FileReader(new File(s)));
          }
  
          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;
          }
      }
  
  }