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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  
  public class TaskC {
      final static long mod = 998244353L;
  
      public static void main(String[] arg) {
          final FastScanner in = new FastScanner(System.in);
          final PrintWriter out = new PrintWriter(System.out);
          final int T = in.nextInt();
          for (int t = 0; t < T; t++) {
              final int n = in.nextInt();
              final int a[] = in.readIntArr(n);
              final long sol = solution(n, a);
              out.println(sol);
          }
          out.flush();
          out.close();
          in.close();
      }
  
      private static long solution(int n, int[] a) {
  
          final int max = max(a);
          int[] cntMap = new int[3];
          for (int i = 0; i < n; i++) {
              if (a[i] == max) {
                  cntMap[2]++;
                  continue;
              }
              if (a[i] + 1 == max) {
                  cntMap[1]++;
                  continue;
              }
              cntMap[0]++;
          }
          if (cntMap[2] >= 2) {
              return fact(n);
          }
          if (cntMap[1] == 0) {
              return 0;
          }
          long res = fact(n);
          long divide = res;
          while (divide % (cntMap[1] + 1) != 0) {
              divide += mod;
          }
          long minusSum = divide / (cntMap[1] + 1);
  //        for (int maxPos = n; maxPos - 1 >= cntMap[1]; maxPos--) {
  //            minusSum += fact(maxPos - 1) * combinations(cntMap[0], n - maxPos);
  //            minusSum%=mod;
  //  }
  
          return (res + mod - minusSum) % mod;
      }
  
      final static long combinations(final long n, final long k) {
          return (fact(n) / fact(n - k)) % mod;
      }
  
      final static int max(final int[] a) {
          int max = a[0];
          for (int i = 1; i < a.length; i++) {
              if (max < a[i]) {
                  max = a[i];
              }
          }
          return max;
      }
  
      final static long fact(final long n) {
          long res = 1;
          for (int i = 2; i <= n; i++) {
              res *= i;
              res = res % mod;
          }
          return res;
      }
  
  
      private static class FastScanner {
          BufferedReader br;
          StringTokenizer st;
  
          FastScanner(InputStream stream) {
              try {
                  br = new BufferedReader(new InputStreamReader(stream));
              } catch (Exception e) {
                  e.printStackTrace();
              }
          }
  
          String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          int[] readIntArr(int n) {
              int[] result = new int[n];
              for (int i = 0; i < n; i++) {
                  result[i] = Integer.parseInt(next());
              }
              return result;
          }
  
          long[] readLongArr(int n) {
              long[] result = new long[n];
              for (int i = 0; i < n; i++) {
                  result[i] = Long.parseLong(next());
              }
              return result;
          }
  
  
          void close() {
              try {
                  br.close();
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
      }
  
  
  }