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: // Compete Against yourself :)
  
  import java.io.*;
  import java.util.*;
  
  public class TaskC {
      static FastReader fr = new FastReader();
      StringBuilder op = new StringBuilder();
      private final int mod = 998_244_353;
  
      public void run() {
          int tt=fr.nextInt();
          while(tt-->0){
              int n=fr.nextInt();
              int a[]=new int[n];
              for(int i=0;i<n;i++){
                  a[i]=fr.nextInt();
              }
              int max=0;
              for(int x:a){
                  max=Math.max(max,x);
              }
              int cnt=0;
              for(int x:a){
                  if(x==max){
                      cnt++;
                  }
              }
              int k=0;
              for(int x:a){
                  if(x==max-1){
                      ++k;
                  }
              }
              int ans=1,sub=1;
              for(int i=1;i<=n;i++){
                  ans = mul(ans,i);
                  if(i!=k+1){
                      sub = mul(sub,i);
                  }
              }
              if(cnt==1){
                  ans = (ans-sub+mod)%mod;
              }
              op.append(ans).append("\n");
          }
          System.out.println(op);
      }
      private int add(int a,int b){
          return (a%mod+b%mod)%mod;
      }
      private int mul(int a,int b){
          long ans=a;
          ans = (ans%mod);
          ans = (ans%mod)*(b%mod)%mod;
          return (int)(ans%mod);
      }
  
      public static void main(String[] args) {
          new TaskC().run();
      }
      //math related functions
  
      //gcd
      private long gcd(long a, long b) {
          if (a == 0) {
              return b;
          } else {
              return gcd(b % a, a);
          }
      }
  
      //lcm
      private long lcm(long a, long b) {
          return (a * b) / gcd(a, b);
      }
  
      //exponential power
      private long power(long a, long b) {
          if (a == 1 || b == 0) {
              return 1;
          }
          long res = 1;
          while (b != 0) {
              if (b % 2 == 1) {
                  res = res * a;
              }
              b /= 2;
              a *= a;
          }
          return res;
      }
  
      // euler totient function copied from
      // https://www.geeksforgeeks.org/eulers-totient-function/
  
      private int phi(int n) {
          // Initialize result as n
          double result = n;
  
          // Consider all prime factors of n and for
          // every prime factor p, multiply result
          // with (1 - 1/p)
          for (int p = 2; p * p <= n; ++p) {
              // Check if p is a prime factor.
              if (n % p == 0) {
                  // If yes, then update n and result
                  while (n % p == 0)
                      n /= p;
                  result *= (1.0 - (1.0 / (double) p));
              }
          }
  
          // If n has a prime factor greater than sqrt(n)
          // (There can be at-most one such prime factor)
          if (n > 1)
              result *= (1.0 - (1.0 / (double) n));
  
          return (int) result;
      }
  
  
      //binary search related functions
      private int lowerBound(int[] a, int key) {
          int ans = -1;
          int low = 0;
          int high = a.length - 1;
          while (low <= high) {
              int mid = (low + high) >>> 1;
              if (a[mid] >= key) {
                  ans = mid;
                  high = mid - 1;
              } else {
                  low = mid + 1;
              }
          }
          return ans;
      }
  
      private int upperBound(int[] a, int key) {
          int ans = -1;
          int low = 0;
          int high = a.length - 1;
          while (low <= high) {
              int mid = (low + high) >>> 1;
              if (a[mid] > key) {
                  ans = mid;
                  high = mid - 1;
              } else {
                  low = mid + 1;
              }
          }
          return ans;
      }
  
      //reverse
      private void reverse(int[] a, int s, int e) {
          for (int i = s, j = e; i <= j; i++, j--) {
              int temp = a[i];
              a[i] = a[j];
              a[j] = temp;
          }
      }
  
      private void reverse(long[] a, int s, int e) {
          for (int i = s, j = e; i <= j; i++, j--) {
              long temp = a[i];
              a[i] = a[j];
              a[j] = temp;
          }
      }
  
      //sorting
      private void sort(int[] a) {
          List<Integer> ls = new ArrayList<Integer>();
          for (int x : a) {
              ls.add(x);
          }
          Collections.sort(ls);
          for (int i = 0; i < a.length; i++) {
              a[i] = ls.get(i);
          }
      }
  
      private void sort(long[] a) {
          List<Long> ls = new ArrayList<Long>();
          for (long x : a) {
              ls.add(x);
          }
          Collections.sort(ls);
          for (int i = 0; i < a.length; i++) {
              a[i] = ls.get(i);
          }
      }
  
      private void sortInRangeInt(int[] a, int s, int e) {
          List<Integer> ls = new ArrayList<>();
          for (int i = s; i <= e; i++) ls.add(a[i]);
          Collections.sort(ls);
          for (int i = s; i <= e; i++) {
              a[i] = ls.get(i);
          }
      }
  
      private void sortInRangeLong(long[] a, int s, int e) {
          List<Long> ls = new ArrayList<>();
          for (int i = s; i <= e; i++) ls.add(a[i]);
          Collections.sort(ls);
          for (int i = s; i <= e; i++) {
              a[i] = ls.get(i);
          }
      }
  
      //Fast I/O
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              br = new BufferedReader(new InputStreamReader(System.in));
              st = null;
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException ie) {
                      ie.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          ;
  
          int[] readIntArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) {
                  a[i] = nextInt();
              }
              return a;
          }
  
          long[] readLongArray(int n) {
              long[] a = new long[n];
              for (int i = 0; i < n; i++) {
                  a[i] = nextLong();
              }
              return a;
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException ie) {
                  ie.printStackTrace();
              }
              return str;
          }
      }
  }
  