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 Main {
  
      static long mod = 998244353;
  
      static long binp(long a, long p) {
          long ans = 1;
          while (p > 0) {
              if (p % 2 == 1)
                  ans = ans * a % mod;
              a = a * a % mod;
              p >>= 1;
          }
          return ans;
      }
  
      static long divide_mod(long a, long b) {
          return a * binp(b, mod - 2) % mod;
      }
  
      static long fact(int a) {
          long x = 1;
          for (int i = 2; i <= a; i++) {
              x = (x * i) % mod;
          }
          return x;
      }
  
      public static void main(String[] args) throws IOException {
          //br = new BufferedReader(new FileReader("input.txt"));
          //out = new PrintWriter("output.txt");
          int t = nextInt();
          while (t-- != 0) {
              int n = nextInt();
              Long[] a = new Long[n];
              long max1 = -1;
              long max2 = -1;
              TreeSet<Long> s = new TreeSet<>();
              TreeMap<Long, Integer> tm = new TreeMap<>();
              for (int i = 0; i < n; i++) {
                  a[i] = nextLong();
                  s.add(a[i]);
                  tm.put(a[i], tm.getOrDefault(a[i], 0) + 1);
                  if (max1 < a[i]) {
                      max2 = max1;
                      max1 = a[i];
                  } else if (max2 < a[i] && a[i] != max1) max2 = a[i];
              }
              if (max2 == -1) out.println(fact(n));
              else {
                  if (tm.get(max1) == 1 && max1 - max2 > 1) out.println(0);
                  else {
                      long ans = 0;
                      if (tm.get(max1) > 1) {
                          ans=fact(n);
                      } else {
                          long z = 1;
                          for (int i = tm.get(max2) + 2; i < n + 1;++i)
                              z = z * i % mod;
                          ans = fact(tm.get(max2)) * tm.get(max2) % mod *z % mod;
                      }
                      out.println(ans);
                  }
              }
          }
          ;
          out.close();
      }
  
      static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter out = new PrintWriter(System.out);
      static StringTokenizer in = new StringTokenizer("");
  
      public static boolean hasNext() throws IOException {
          if (in.hasMoreTokens()) return true;
          String s;
          while ((s = br.readLine()) != null) {
              in = new StringTokenizer(s);
              if (in.hasMoreTokens()) return true;
          }
          return false;
      }
  
      public static String nextToken() throws IOException {
          while (!in.hasMoreTokens()) {
              in = new StringTokenizer(br.readLine());
          }
          return in.nextToken();
      }
  
      public static int nextInt() throws IOException {
          return Integer.parseInt(nextToken());
      }
  
      public static long nextLong() throws IOException {
          return Long.parseLong(nextToken());
      }
  
      public static double nextDouble() throws IOException {
          return Double.parseDouble(nextToken());
      }
  }
  
  class Ayo implements Comparable<Ayo> {
      long a;
      long b;
  
      Ayo(long a, long b) {
          this.a = a;
          this.b = b;
      }
  
      public int compareTo(Ayo o) {
          if (o.a == a) return -Long.compare(b, o.b);
          else return Long.compare(a, o.a);
      }
  }