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: #include <bits/stdc++.h>
  #pragma GCC optimize("O3")
  #pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
  using namespace std;
  template <typename val1>
  void __f(const char* name, val1&& value) {
    cout << name << " = " << value << endl;
  }
  template <typename val1, typename... values>
  void __f(const char* names, val1&& value, values&&... multiplevalue) {
    const char* comma = strchr(names + 1, ',');
    cout.write(names, comma - names) << " = " << value << " ";
    __f(comma + 1, multiplevalue...);
  }
  const long long N = 2e5 + 5;
  vector<long long> fac(N, 1);
  long long power(long long x, int y, int p) {
    long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long modInverse(long long n, int p) { return power(n, p - 2, p); }
  long long nCr(long long n, long long r, long long p) {
    if (n < r) return 0;
    if (r == 0) return 1;
    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
           p;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    const long long mod = 998244353;
    for (long long i = 2; i <= 2e5; i++) fac[i] = (fac[i - 1] * i) % mod;
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      int arr[n];
      for (int i = 0; i < n; i++) cin >> arr[i];
      sort(arr, arr + n);
      map<int, int> ok;
      for (int i = 0; i < n; i++) ok[arr[i]]++;
      if (ok[arr[n - 1]] > 1) {
        cout << fac[n] << endl;
      } else if (ok[arr[n - 1] - 1] == 0) {
        cout << 0 << endl;
      } else {
        long long minus = 0;
        for (int i = 0; i < n; i++) {
          int left = n - ok[arr[n - 1] - 1] - 1;
          int have = ok[arr[n - 1] - 1];
          if (i < have) continue;
          minus =
              minus + (((nCr(i, have, mod) * fac[have]) % mod) * fac[left]) % mod;
          minus %= mod;
        }
        cout << (fac[n] - minus + mod) % mod << endl;
      }
    }
    cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
         << "\n";
    return 0;
  }
  