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 dbg() { cout << "DEBUG\n"; }
  const int INF = 1e9;
  const long long LINF = 1e18;
  const long long mod = 998244353;
  const int maxn = 2e5 + 100;
  long long fac[maxn];
  long long inv(int x) {
    long long second = 1;
    for (; x > 1; x = mod % x) second = second * (mod - mod / x) % mod;
    return second;
  }
  void pre() {
    fac[0] = 1;
    for (int i = 1; i <= 200010; i++) fac[i] = (fac[i - 1] * i) % mod;
    return;
  }
  long long C(int n, int m) {
    if (m > n) return 0;
    if (m == 0) return 1;
    return (fac[n] * inv(fac[m])) % mod * inv(fac[n - m]) % mod;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<int> arr(n + 1);
    for (int i = 1; i <= n; i++) cin >> arr[i];
    int mx = -1, sc_mx = -1;
    for (int i = 1; i <= n; i++) {
      if (arr[i] > mx) {
        sc_mx = mx;
        mx = arr[i];
      } else if (arr[i] > sc_mx)
        sc_mx = arr[i];
    }
    if (mx - sc_mx > 1) {
      cout << "0" << '\n';
      return;
    }
    int cnt_mx = 0, cnt_se_mx = 0;
    for (int i = 1; i <= n; i++) {
      if (arr[i] == mx)
        cnt_mx++;
      else if (arr[i] == sc_mx)
        cnt_se_mx++;
    }
    long long tmp = (((1 + n - 1) * (n - 1) % mod) * inv(2) % mod);
    if (cnt_mx > 1) {
      long long ans = fac[n];
      cout << ans << '\n';
    } else if (cnt_se_mx > 1) {
      long long ans = fac[n];
      for (int i = n - 1; i >= cnt_se_mx; i--) {
        ans -= (C(n - cnt_se_mx - 1, i - cnt_se_mx) * fac[i] % mod) *
               fac[n - i - 1] % mod;
        if (ans < 0) ans += mod;
      }
      cout << ans << '\n';
    } else {
      long long ans = (tmp * fac[n - 2]) % mod;
      cout << ans << '\n';
    }
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    pre();
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) solve();
    return 0;
  }
  