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(vector<long long> v) {
    cout << "[ ";
    for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
    cout << "]"
         << "\n";
  }
  long long Power(long long $a, long long $b) {
    long long $result = 1;
    $a %= 998244353ll;
    while ($b) {
      if ($b % 2 == 1) $result *= $a;
      $a *= $a;
      $a %= 998244353ll;
      $result %= 998244353ll;
      $b /= 2;
    }
    return $result;
  }
  long long Mod_Inv(long long $a) {
    $a %= 998244353ll;
    $a = Power($a, 998244353ll - 2);
    return $a;
  }
  long long Factorial[500001];
  long long Make_Factorial() {
    Factorial[0] = 1;
    for (long long i = 1; i < 500001; i++) {
      Factorial[i] = i * Factorial[i - 1];
      Factorial[i] %= 998244353ll;
    }
    return 0;
  }
  long long Implement_Make_Factorial = Make_Factorial();
  long long nCr(long long $n, long long $r) {
    if ($n < $r) return 0;
    long long $ans = Factorial[$n];
    $ans %= 998244353ll;
    $ans *= Mod_Inv(Factorial[$r]);
    $ans %= 998244353ll;
    $ans *= Mod_Inv(Factorial[$n - $r]);
    $ans %= 998244353ll;
    return $ans;
  }
  long long nPr(long long $n, long long $r) {
    long long $ans = Factorial[$n];
    $ans %= 998244353ll;
    $ans *= Mod_Inv(Factorial[$n - $r]);
    $ans %= 998244353ll;
    return $ans;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> arr(n);
    set<long long> s;
    map<long long, long long> mp;
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
      s.insert(arr[i]);
      if (mp.find(arr[i]) == mp.end())
        mp[arr[i]] = 1;
      else
        mp[arr[i]]++;
    }
    sort(arr.begin(), arr.end());
    if (arr[n - 1] - arr[n - 2] >= 2 && mp[arr[n - 1]] == 1) {
      cout << 0 << "\n";
      return;
    }
    long long ans = Factorial[n];
    if (mp[arr[n - 1]] > 1) {
      cout << ans << "\n";
      return;
    }
    long long cnt = mp[arr[n - 2]] + 1;
    ans *= Mod_Inv(Factorial[cnt]);
    ans %= 998244353ll;
    ans *= Factorial[cnt - 1];
    ans %= 998244353ll;
    long long anss =
        (Factorial[n] % 998244353ll - ans % 998244353ll) % 998244353ll;
    if (anss < 0) anss += 998244353ll;
    cout << anss << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  