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(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(double t) { cerr << t; }
  long long int powe(long long int x, long long int y,
                     long long int p = 998244353) {
    long long int 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 int inverse_modulo(long long int a, long long int p = 998244353) {
    return powe(a, p - 2, p);
  }
  vector<long long int> fact(200001, 1);
  void fill_fact(long long int p = 998244353) {
    for (int i = 1; i < (int)200001; i++) {
      fact[i] = (fact[i - 1] * i) % p;
    }
  }
  template <class T, class vv>
  void _print(pair<T, vv> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class vv>
  void _print(map<T, vv> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class vv>
  void _print(pair<T, vv> 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 vv>
  void _print(map<T, vv> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    int ma = 0, ma2 = 0;
    for (int i = 0; i < (int)n; i++) {
      cin >> a[i];
      ma = max(ma, a[i]);
    }
    int cnt1 = 0, cnt2 = 0;
    for (int i = 0; i < (int)n; i++) {
      if (a[i] != ma)
        ma2 = max(ma2, a[i]);
      else
        cnt1++;
    }
    if (cnt1 > 1) {
      cout << fact[n] << "\n";
      return;
    }
    if (ma - ma2 > 1) {
      cout << 0 << "\n";
      return;
    }
    for (int i = 0; i < (int)n; i++) {
      if (a[i] == ma2) cnt2++;
    }
    long long int ans = ((cnt2 * fact[n]) % 998244353) * inverse_modulo(cnt2 + 1);
    ans %= 998244353;
    cout << ans << "\n";
  }
  int main() {
    freopen("Error.txt", "w", stderr);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    fill_fact();
    int tt = 1;
    cin >> tt;
    while (tt--) {
      solve();
    }
    return 0;
  }
  