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.io.*;
  public class c{
    public static void main(String[] args_are_sad_whyyyyyy) throws Exception {
      BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));PrintWriter writer=new PrintWriter(System.out);
      int T = Integer.parseInt(reader.readLine()); while (T --> 0) {
        int n = Integer.parseInt(reader.readLine());
        List<Integer> a = new ArrayList<>();
        StringTokenizer tokens = new StringTokenizer(reader.readLine());
        for (int i = 0; i < n; i++) {
          a.add(Integer.parseInt(tokens.nextToken()));
        } Collections.sort(a);
        writer.println(solve(a));
      }
      writer.close();
    }
    static List<Integer> defenestrations = out_the_window(200001, 998244353);
    static List<Integer> inverse = in_the_window(defenestrations, 998244353);
    static List<Integer> out_the_window(int n, int mod) {
      List<Integer> out = new ArrayList<>();
      out.add(1); long x = 1;
      for (int i = 1; i < n; i++) {
        x *= i; x %= mod; out.add((int)x); 
      } return out;
    }
    static List<Integer> in_the_window(List<Integer> f, int mod) {
      List<Integer> out = new ArrayList<>();
      for (int it : f) {
        long res = 1;
        for (int e = mod - 2; e > 0; e /= 2) {
          if (e % 2 == 1) res = res * it % mod;
          it = (int)((long)it * it % mod);
        }
        out.add((int)res);
      }
      return out;
    }
    static int mod = 998244353;
    static int solve(List<Integer> cactus) {
      int largest = cactus.get(cactus.size() - 1);
      int secondLargest = cactus.get(cactus.size() - 2);
      if (largest == secondLargest) return defenestrations.get(cactus.size());
      int kaktus = cactus.size();
      if (largest == secondLargest + 1) {
        int numSecond = (int)cactus.stream().filter(x -> x == secondLargest).count();
        long bad = 0;
        for (int i = 0; i < kaktus; i++) {
          bad += choose(kaktus - 1 - numSecond, i) * defenestrations.get(i) % 998244353 *
              defenestrations.get(kaktus - 1 - i) % 998244353;
        }
        return (int)((defenestrations.get(kaktus) - bad) % mod + mod) % mod;
      }
      return 0;
    }
    static long choose(int n, int r) {
      if (r < 0 || n < r) return 0;
      return (long)defenestrations.get(n) * inverse.get(r) % mod * inverse.get(n - r) % mod;
    }
  }