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;
  template <typename A>
  ostream &operator<<(ostream &cout, vector<A> const &v);
  template <typename A, typename B>
  ostream &operator<<(ostream &cout, pair<A, B> const &p) {
    return cout << "(" << p.first << ", " << p.second << ")";
  }
  template <typename A>
  ostream &operator<<(ostream &cout, vector<A> const &v) {
    cout << "[";
    for (int i = 0; i < v.size(); i++) {
      if (i) cout << ", ";
      cout << v[i];
    }
    return cout << "]";
  }
  template <typename A, typename B>
  istream &operator>>(istream &cin, pair<A, B> &p) {
    cin >> p.first;
    return cin >> p.second;
  }
  mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());
  void usaco(string filename) {
    freopen((filename + ".in").c_str(), "r", stdin);
    freopen((filename + ".out").c_str(), "w", stdout);
  }
  const long double pi = 3.14159265358979323846;
  const long long mod = 998244353;
  long long n, m, q, k, l, r, x, y, z;
  const long long template_array_size = 1e6 + 17646;
  long long a[template_array_size];
  long long b[template_array_size];
  long long c[template_array_size];
  string second, t;
  long long fact[200005];
  long long cnt(long long x) {
    long long r = 0;
    for (long long i = 0; i < n; i++) {
      if (a[i] == x) {
        ++r;
      }
    }
    return r;
  }
  void solve(int tc = 0) {
    cin >> n;
    for (long long i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    long long y = cnt(a[n - 1]);
    long long z = cnt(a[n - 1] - 1);
    if (y >= 2) {
      cout << fact[n] << '\n';
      return;
    }
    long long r = 1;
    for (long long i = n; i > (y + z); i--) r = (r * i) % mod;
    r = (r * (fact[y + z] - fact[y + z - 1] + mod)) % mod;
    cout << r << '\n';
  }
  int main() {
    { ios_base::sync_with_stdio(false); }
    { cin.tie(NULL); }
    cout << setprecision(15) << fixed;
    fact[0] = fact[1] = 1;
    for (long long i = 2; i <= 200003; i++) {
      fact[i] = (i * fact[i - 1]) % mod;
    }
    int tc = 1;
    cin >> tc;
    for (int t = 0; t < tc; t++) {
      solve(t);
    }
  }
  