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;
  void _print(long long int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(vector<vector<T>> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " , ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(vector<vector<T>> v) {
    cerr << "[ ";
    for (auto& row : v) {
      cerr << " {";
      for (auto& col : row) {
        _print(col);
        cerr << " , ";
      }
      cerr << "}\n";
    }
    cerr << " ]";
  }
  const long long int no = 2e5 + 5, modulo = 998244353, inf = 1e18, N = 2e3 + 9;
  long long int ar[no], br[no], cr[no];
  long long int calc(long long int n) {
    if (n == 0) return 1;
    return (n * calc(n - 1)) % modulo;
  }
  long long int calc1(long long int n, long long int x) {
    if (n == 0) return 1;
    if (n == x)
      return calc1(n - 1, x) % modulo;
    else
      return (n * calc1(n - 1, x)) % modulo;
  }
  void solve() {
    long long int n = 0, m = 0, a = 0, b = 0, c = 0, d = 0, x = 0, y = 0, z = 0,
                  w = 0, k = 0;
    cin >> n;
    for (long long int i = 0; i < n; i += 1) cin >> ar[i];
    sort(ar, ar + n);
    if (ar[n - 1] - ar[n - 2] > 1) {
      cout << 0 << "\n";
      return;
    }
    if (ar[n - 1] == ar[n - 2]) {
      cout << calc(n) << "\n";
      return;
    }
    for (long long int i = 0; i < n; i += 1) {
      if (ar[n - 2] == ar[i]) {
        x = i;
        break;
      }
    }
    y = calc1(n, n - x);
    cout << (y * (n - x - 1)) % modulo << "\n";
  }
  inline void runn() {}
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    runn();
    long long int t = 1;
    cout << fixed << setprecision(15);
    cin >> t;
    for (long long int i = 1; i < t + 1; i += 1) {
      solve();
    }
  }
  