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 C {
      //--------------------------INPUT READER--------------------------------//
      static class fs {
          public BufferedReader br;
          StringTokenizer st = new StringTokenizer("");
  
          public fs() { this(System.in); }
          public fs(InputStream is) {
              br = new BufferedReader(new InputStreamReader(is));
          }
          String next() {
              while (!st.hasMoreTokens()) {
                  try { st = new StringTokenizer(br.readLine()); }
                  catch (IOException e) { e.printStackTrace(); }
              }
              return st.nextToken();
          }
  
          int ni() { return Integer.parseInt(next()); }
          long nl() { return Long.parseLong(next()); }
          double nd() { return Double.parseDouble(next()); }
          String ns() { return next(); }
  
          int[] na(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = ni();
              return a;
          }
      }
      //-----------------------------------------------------------------------//
  
      //---------------------------PRINTER-------------------------------------//
      static class Printer {
          static PrintWriter w;
          public Printer() {this(System.out);}
          public Printer(OutputStream os) {
              w = new PrintWriter(os);
          }
          public void p(int i) {w.println(i);};
          public void p(long l) {w.println(l);};
          public void p(double d) {w.println(d);};
          public void p(String s) { w.println(s);};
          public void pr(int i) {w.println(i);};
          public void pr(long l) {w.print(l);};
          public void pr(double d) {w.print(d);};
          public void pr(String s) { w.print(s);};
          public void pl() {w.println();};
          public void close() {w.close();};
      }
      //------------------------------------------------------------------------//
  
      //--------------------------VARIABLES------------------------------------//
      static fs sc = new fs();
      static OutputStream outputStream = System.out;
      static Printer w = new Printer(outputStream);
      static long mod = 998_244_353;
      static long[] fact;
      //-----------------------------------------------------------------------//
  
      //--------------------------ADMIN_MODE-----------------------------------//
      private static void ADMIN_MODE() throws IOException {
          if (System.getProperty("ONLINE_JUDGE") == null) {
              w = new Printer(new FileOutputStream("output.txt"));
              sc = new fs(new FileInputStream("input.txt"));
          }
      }
      //-----------------------------------------------------------------------//
  
      //----------------------------START--------------------------------------//
      static long tc = 0;
      public static void main(String[] args)
              throws IOException {
  
          ADMIN_MODE();
          fact = new long[1000000];
          fact[0] = 0;
          fact[1] = 1;
          for(int i = 2; i < 1000000; i++) {
              fact[i] = fact[i-1]*i;
              fact[i] %= mod;
          }
  
          int t = sc.ni();
          while(t-->0) {
              tc++;
              solve();
          }
  
  
          w.close();
      }
  
      static void solve() throws IOException {
          int n = sc.ni();
          Long[] arr = new Long[n];
          for(int i = 0; i < n; i++) {
              arr[i] = sc.nl();
          }
          Arrays.sort(arr);
          long pre = arr[0];
          List<Long> count = new ArrayList<>();
          count.add(1L);
          for(int i = 1; i < n; i++) {
              if(arr[i] == pre) {
                  count.set(count.size()-1, count.get(count.size()-1)+1);
              } else {
                  pre = arr[i];
                  count.add(1L);
              }
          }
          long ans = 1;
          ans = fact[n];
  
          int nn = count.size();
          if(count.size() == 1) {
              w.p(ans);
              return;
          }
  
          long prepre = 0;
          long target = arr[n-1];
          for(int i = n-1; i >= 0; i--) {
              if(arr[i] != target) {
                  prepre = arr[i];
                  break;
              }
          }
          if(arr[n-1].equals(arr[n-2])) {
              w.p(ans);
              return;
          }
  
          if(target-prepre > 1) {
              w.p(0);
              return;
          }
  
  
          long extra = 0;
          long st = count.get(nn-2);
          for(long i = st; i <= n-1; i++) {
              long curr1 = fact[(int)i];
              if(n-st-1 >= 1 && i-st >= 1 && n-st-1 > i-st && i != n-1) {
                  curr1 *= (ncr(n-st-1, i - st));
                  curr1 %= mod;
              }
              if(n-i-1 > 0) {
                  curr1 *= fact[(int) (n-i-1)];
                  curr1 %= mod;
              }
              curr1 %= mod;
              extra += curr1;
              extra %= mod;
          }
  
          w.p((ans-extra+mod)%mod);
      }
      
      static long fastP(long b, long x, long mod) {
          if (x == 0) return 1;
          if (x == 1) return b;
          if (x % 2 == 0) return fastP(b * b % mod, x / 2, mod) % mod;
  
          return b * fastP(b * b % mod, x / 2, mod) % mod;
      }
  
      static long ncr(long n, long r) {
          long num = fact[(int)n];
          long den = fact[(int)(n-r)];
          den%=mod;
          den *= fact[(int)(r)];
          den%=mod;
          return ((num%mod)*fastP((den%mod),mod-2, mod))%mod;
      }
  }