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.*;
  
  public class tank {
  
      static final FastScanner fs = new FastScanner();
  
      static PrintWriter out = new PrintWriter(System.out);
  
      public static void main(String[] args) {
          int t = fs.nextInt();
          while(t-- > 0) run_case();
  
          out.close();
      }
  
      static void run_case() {
          int n = fs.nextInt();
          int[] a = fs.readArray(n);
  
          HashMap<Integer, Integer> hm = new HashMap<>();
  
          ruffleSort(a);
  
          boolean eq = true;
  
          int mx = a[n - 1], smx = a[0];
  
          hm.put(a[0], hm.getOrDefault(a[0], 0) + 1);
  
          for (int i = 1; i < n; i++) {
              if(a[i] < mx) smx = Math.max(smx, a[i]);
  
              hm.put(a[i], hm.getOrDefault(a[i], 0) + 1);
          }
  
          if(hm.get(mx) > 1) {
  
          }
          else {
              if(a[n - 1] - a[n - 2] > 1) {
                  out.println(0);
                  return;
              }
              eq = false;
          }
  
          long mod = 998244353, nf = 1;
  
          long[] pref = new long[n + 1];
  
          pref[0] = 1;
  
          for (int i = 1; i <= n; i++) {
              nf = (nf * i) % mod;
              pref[i] = nf;
          }
  
          if(eq) {
              out.println(nf);
          }else {
              int red = n - 1 - hm.getOrDefault(smx, 0);
              long ans = 0, tt = red - 1;
  
              long[] fkthis = new long[red + 1];
              fkthis[0] = 1;
  
              for (int i = 1; i <= red; i++) {
                  if(i == 1) fkthis[1] = red;
                  else {
                      fkthis[i] = (fkthis[i - 1] * (red - i + 1)) % mod;
                  }
              }
  
              for (int i = 0; i <= red; i++) {
                  ans = (ans + (pref[n - 1 - i] * fkthis[i]) % mod) % mod;
              }
  
              out.println((nf - ans + mod) % mod);
          }
      }
  
      static void ruffleSort(int[] a) {
          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;
          }
  
          Arrays.sort(a);
      }
  
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          String nextLine(){
              try {
                  return br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
  
              return "";
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
      }
  }