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.InputStreamReader;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.List;
  import java.util.StringTokenizer;
  
  public class c {
      private static final int MAX = (int) 2e5 + 10;
      private static final int mod = 998244353;
  
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          int tc = Integer.parseInt(br.readLine());
  
          long[] factorial = new long[MAX];
          factorial[0] = 1;
          for (int i = 1; i < MAX; i++) {
              factorial[i] = factorial[i - 1] * i;
              factorial[i] %= mod;
          }
  
          for (int tt = 0; tt < tc; tt++) {
              int n = Integer.parseInt(br.readLine());
              st = new StringTokenizer(br.readLine());
              List<Integer> list = new ArrayList<>();
              while (st.hasMoreTokens()) list.add(Integer.parseInt(st.nextToken()));
  
              list.sort(Integer::compareTo);
              int max = list.get(n - 1);
              int maxSecond = list.get(n - 2);
  
              int[] cnt = new int[2]; // 0 : max , 1 : secondMax
              for (int i = 0; i < n; i++) {
                  if (list.get(i) == max) cnt[0]++;
                  if (list.get(i) == maxSecond) cnt[1]++;
              }
  
  //            최대값과 2번째 최대값이
  //             0 차이,  최대값이 2개 이상 : 어케든 ㅇㅋ
  //            1 차이 : 젤 큰거 뒤에 1차이 나는 거 둬야 함 ==젤 큰 거 뒤에 1차이 나는게 없는 상황 빼줘야 함
  //            2 이상 차이 : 절대 안댐 ?
  
              long ans = factorial[n];
  
              if (max - maxSecond == 0) {
                  ans = factorial[n];
  
              } else if (max - maxSecond == 1) {
  
                  int othersCnt = n - 1 - cnt[1];
  
                  for (int f = 0; f <= othersCnt; f++) {
                      long subtract = factorial[cnt[1] + f];
  
                      long A = factorial[othersCnt];
                      long B = (factorial[othersCnt - f] * factorial[f]) % mod;
  
                      long inverse = fastPow(B, mod - 2);
                      subtract *= ((A * inverse % mod) % mod);
                      subtract %= mod;
                      
                      subtract *= factorial[othersCnt - f];
                      subtract %= mod;
  
  //                    System.out.println(B+" "+inverse+" "+(A*inverse));
  //                    System.out.println(factorial[cnt[1]+f]+" "+((A * inverse) % mod)+" "+factorial[othersCnt - f]);
                      
                      ans+=mod;
                      ans -= subtract;
                      ans %= mod;
  //                    System.out.println(subtract);
                  }
              } else {
                  ans = 0;
              }
  
              if (cnt[0] >= 2) {
                  ans = factorial[n];
              }
  
              System.out.println(ans);
          }
      }
  
  
      private static long fastPow(long x, long e) {
          if (e == 0) return 1;
          if (e % 2 == 1) return (fastPow(x, e - 1) * x) % mod;
          else {
              long half = fastPow(x, e / 2) % mod;
              return (half * half) % mod;
          }
      }
  
  }