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;
  long long f[200005];
  int a[200005], n;
  long long getmod(long long x) {
    long long res = 1;
    int t = 998244353 - 2;
    for (; t; t /= 2, x = (x * x) % 998244353)
      if (t & 1) res = (res * x) % 998244353;
    return res;
  }
  long long C(int k, int n) {
    long long res = f[n];
    res = (res * getmod(f[k])) % 998244353;
    res = (res * getmod(f[n - k])) % 998244353;
    return res;
  }
  void Solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    sort(a + 1, a + n + 1, greater<int>());
    if (a[1] == a[2]) {
      cout << f[n] << '\n';
      return;
    }
    if (a[1] - a[2] > 1) {
      cout << 0 << '\n';
      return;
    }
    int cnt = 0;
    for (int i = 2; i <= n; ++i)
      if (a[i] == a[2]) ++cnt;
    long long ans = 0;
    for (int i = 1; i < n; ++i) {
      if (cnt >= i)
        ans = (ans + f[n - 1]) % 998244353;
      else {
        ans = (ans + (f[n - 1] -
                      ((f[cnt] * C(cnt, i - 1) % 998244353) * f[n - cnt - 1] %
                       998244353) +
                      998244353)) %
              998244353;
      }
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(NULL);
    cin.tie(NULL);
    cout.tie(NULL);
    if (fopen("D:\\C++\\clion\\"
              ".inp",
              "r")) {
      freopen(
          "D:\\C++\\clion\\"
          ".inp",
          "r", stdin);
      freopen(
          "D:\\C++\\clion\\"
          ".out",
          "w", stdout);
    }
    f[0] = 1;
    for (int i = 1; i < 200005; ++i) f[i] = f[i - 1] * i % 998244353;
    int t;
    cin >> t;
    while (t--) {
      Solve();
    }
    cerr << "Time collapse : " << fixed << setprecision(3)
         << 1.000 * clock() / CLOCKS_PER_SEC;
    return 0;
  }
  