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.io.PrintWriter;
  import java.util.*;
  
  
  public class Main {
      static int i, j, k, n, m, t, y, x, sum = 0;
      static long mod =  998244353;
      static FastScanner fs = new FastScanner();
      static PrintWriter out = new PrintWriter(System.out);
      static String str;
      static long ans;
  
  
      public static void main(String[] args) {
  
          t = fs.nextInt();
          long[] fact = new long[200005];
          fact[0]=1L;
  
          for(i=1;i<200005;i++){
              fact[i] = fact[i-1]*i;
              fact[i]%=mod;
  
          }
  
          while(t-- >0) {
  
              n = fs.nextInt();
  
              int[] arr = new int[n];
              int max=0, nextMax=0, maxCount = 0, nextMaxCount=0;
  
              for(i=0;i<n;i++){
                  arr[i]=fs.nextInt();
                  max = Math.max(max,arr[i]);
              }
  
              for(i=0;i<n;i++){
                  if(arr[i]!=max)
                      nextMax = Math.max(nextMax,arr[i]);
              }
  
              for(i=0;i<n;i++){
                  if(arr[i] == max)
                      maxCount++;
                  else if(arr[i] == nextMax)
                      nextMaxCount++;
              }
  
              if(maxCount ==1 && (nextMax == 0 || nextMax!=max-1 ))
                  out.println(0);
  
              else{
                  if(maxCount>1)
                      out.println(fact[n]);
  
                  else{
  
                      int x = n-(maxCount+nextMaxCount);
                      ans = fact[n];
  
                      long diff = 1;
                      int j=n-1;
                      while(x>=0){
  
  
                          diff%=mod;
  
                          long temp = (diff*fact[j--])%mod;
  
                          ans-=temp;
  
                          if(ans<0)
                              ans+=mod;
  
                          ans%=mod;
  
                          diff*=(x--);
  
                      }
  
                      out.println(ans);
                  }
  
              }
  
  
  
          }
  
          out.close();
      }
  
      static long computePow(int x, int y){
          long ans = 1;
          for(i=0;i<y;i++){
              ans*=x;
              ans%=mod;
          }
  
          return ans;
      }
  
  
      static long gcd(long a, long b)
      {
          if (b == 0)
              return a;
          return gcd(b, a % b);
      }
  
      static void ruffleSort(int[] a) {
          //ruffle
          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;
          }
  
          //then sort
          Arrays.sort(a);
      }
  
      static void ruffleSort(long[] a) {
          //ruffle
          int n = a.length;
          Random r = new Random();
          for (int i = 0; i < a.length; i++) {
              int oi = r.nextInt(n);
              long 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();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
      static class Pair {
          public int x, y;
  
          Pair(int x, int y){
              this.x = x;
              this.y = y;
          }
      }
  
  
  }