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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.TreeMap;
  import java.util.StringTokenizer;
  import java.util.Map;
  import java.util.Map.Entry;
  import java.io.BufferedReader;
  import java.io.FileReader;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   *
   * @author Khater
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          Scanner in = new Scanner(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          CJuryMeeting solver = new CJuryMeeting();
          solver.solve(1, in, out);
          out.close();
      }
  
      static class CJuryMeeting {
          long[] fac;
  
          long sub(long a, long b, long mod) {
              return (a - b + mod) % mod;
          }
  
          public void solve(int testNumber, Scanner sc, PrintWriter pw) {
  
              fac = new long[(int) (2e5 + 1)];
              long mod = 998244353;
              for (int i = 0; i < fac.length; i++) {
                  if (i == 0) fac[i] = 1;
                  else fac[i] = (fac[i - 1] * i) % mod;
              }
  //		pw.println(Arrays.toString(fac));
  
              int t = 1;
              t = sc.nextInt();
              while (t-- > 0) {
                  int n = sc.nextInt();
                  int[] arr = sc.nextIntArr(n);
                  TreeMap<Integer, Integer> tm = new TreeMap<>();
                  for (int x : arr) tm.put(x, tm.getOrDefault(x, 0) + 1);
                  long ans = fac[n];
                  int max = tm.lastKey();
                  int cnt1 = tm.pollLastEntry().getValue();
                  Integer smax = tm.size() > 0 ? tm.lastKey() : null;
                  if (smax == null || cnt1 > 1) {
                      pw.println(ans);
                  } else {
                      int smaxx = smax.intValue();
                      int cnt2 = tm.pollLastEntry().getValue();
                      if (smaxx != max - 1 && cnt1 == 1) {
                          pw.println(0);
                      } else {
                          int rem = n - cnt1 - cnt2;
                          int x = rem;
                          for (int i = cnt2 + cnt1; i < n; i++) {
  //						pw.println(i+" "+rem+" "+x+" " +(fac[i-1]));
  //						pw.println(ans+" "+((fac[rem]*nCrModPFermat(x,rem,mod))%mod*(fac[i-1]))%mod);
                              ans = sub(ans, (((((fac[rem] * cnt1) % mod) * nCrModPFermat(x, rem, mod)) % mod) * (fac[i - 1])) % mod, mod);
                              rem--;
                          }
                          ans = sub(ans, (fac[n - 1] * cnt1) % mod, mod);
                          pw.println(ans);
                      }
  
                  }
              }
  
          }
  
          long power(long x, long y, long p) {
  
              // Initialize result
              long res = 1;
  
              // Update x if it is more than or
              // equal to p
              x = x % p;
  
              while (y > 0) {
  
                  // If y is odd, multiply x
                  // with result
                  if (y % 2 == 1)
                      res = (res * x) % p;
  
                  // y must be even now
                  y = y >> 1; // y = y/2
                  x = (x * x) % p;
              }
  
              return res;
          }
  
          long modInverse(long n, long p) {
              return power(n, p - 2, p);
          }
  
          long nCrModPFermat(int n, int r,
                             long p) {
  
              if (n < r)
                  return 0;
              // Base case
              if (r == 0)
                  return 1;
  
              // Fill factorial array so that we
              // can find all factorial of r, n
              // and n-r
              return (fac[n] * modInverse(fac[r], p)
                      % p * modInverse(fac[n - r], p)
                      % p)
                      % p;
          }
  
      }
  
      static class Scanner {
          StringTokenizer st;
          BufferedReader br;
  
          public Scanner(FileReader r) {
              br = new BufferedReader(r);
          }
  
          public Scanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
  
          public String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return st.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public int[] nextIntArr(int n) {
              try {
                  int[] a = new int[n];
                  for (int i = 0; i < n; i++)
                      a[i] = nextInt();
                  return a;
              } catch (Exception e) {
                  throw new RuntimeException(e);
              }
  
          }
  
      }
  }
  
  