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>
  using namespace std;
  long long int power(long long int x, long long int b, long long int p = 1) {
    x %= 998244353;
    while (b > 0) {
      if (b & 1) p = (p * x) % 998244353;
      b >>= 1;
      x = (x * x) % 998244353;
    }
    return p;
  }
  vector<long long int> fact;
  void fact_pre(long long int n = 1100000) {
    fact.resize(n, 1);
    for (long long int i = 2; i < n; i++) fact[i] = (fact[i - 1] * i) % 998244353;
  }
  long long int ncr(long long int n, long long int r) {
    if (r > n) return 0;
    return (fact[n] * power((fact[n - r] * fact[r]) % 998244353, 998244353 - 2)) %
           998244353;
  }
  mt19937_64 rang(
      chrono::high_resolution_clock::now().time_since_epoch().count());
  long long int rng(long long int lim, long long int s = 0) {
    uniform_int_distribution<long long int> uid(s, lim - 1);
    return uid(rang);
  }
  double sqr(double n) {
    double s = 0, e = n;
    for (long long int i = 0; i < 50; i++) {
      double m = s + (e - s) / 2;
      if (m * m <= n)
        s = m;
      else
        e = m;
    }
    return s;
  }
  void pre_input() { fact_pre(); }
  void solve() {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a.begin(), a.end());
    reverse(a.begin(), a.end());
    if (a[0] > a[1] + 1) {
      cout << 0 << "\n";
      return;
    }
    if (a[0] == a[1]) {
      cout << fact[n] << "\n";
      return;
    }
    long long int h = 0;
    for (auto &t : a)
      if (t == a[1]) h++;
    long long int k = n - h - 1;
    long long int ans = fact[n];
    for (long long int i = k; i >= 0; i--) {
      ans -= (((ncr(k, i) * fact[i]) % 998244353) * fact[n - i - 1]) % 998244353;
      ans = (ans + 998244353) % 998244353;
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    pre_input();
    long long int tc;
    cin >> tc;
    for (long long int number = 1; number <= tc; number++) {
      solve();
    }
    return 0;
  }
  