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 comment(linker, "/STACK:1000000000,1000000000")
  using namespace std;
  template <class T>
  istream& operator>>(istream& is, vector<T>& vec) {
    for (int i = 0; i < (int)vec.size(); ++i) is >> vec[i];
    return is;
  }
  template <class T>
  ostream& operator<<(ostream& os, vector<T>& vec) {
    for (int i = 0; i < (int)vec.size(); ++i) os << vec[i] << " ";
    return os;
  }
  template <class T>
  istream& operator>>(istream& is, pair<T, T>& a) {
    is >> a.first >> a.second;
    return is;
  }
  const double pi = acos(-1.0);
  const long long mod = 998244353;
  const long long INF = 2e18;
  const int inf = (int)1e9 + 7;
  const int maxlen = (int)1e5 + 100;
  const double eps = 1e-7;
  long long binpow(long long a, long long n) {
    int res = 1;
    while (n) {
      if (n & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      n >>= 1;
    }
    return res;
  }
  long long del(long long a, long long b) {
    long long d = binpow(b, mod - 2);
    return (a * d) % mod;
  }
  void solve() {
    int n;
    cin >> n;
    int mx = -1, mx_n, mn_n = -1;
    for (int i = 0; i < n; i++) {
      int a;
      cin >> a;
      if (a > mx) {
        if (a == mx + 1) {
          mn_n = mx_n;
        } else
          mn_n = 0;
        mx = a;
        mx_n = 0;
      }
      if (a == mx) mx_n++;
      if (a == mx - 1) mn_n++;
    }
    if (mx_n == 1 && mn_n == 0) {
      cout << 0 << endl;
      return;
    }
    if (mx_n > 1) {
      long long fact = 1;
      for (int i = 2; i <= n; i++) {
        fact = (fact * i) % mod;
      }
      cout << fact << endl;
      return;
    }
    long long fact = 1;
    vector<long long> f(n + 1);
    f[0] = 1;
    for (int i = 1; i <= n; i++) {
      fact = (fact * i) % mod;
      f[i] = fact;
    }
    long long sum = 0;
    for (int i = mn_n; i <= n - 1; i++) {
      long long jopa = del(f[i], f[i - mn_n]);
      jopa = (jopa * f[n - mn_n - 1]) % mod;
      sum = (sum + jopa) % mod;
    }
    sum = (f[n] + mod - sum) % mod;
    cout << sum << endl;
  }
  int main() {
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  