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;
  const long long N = 1e6 + 20;
  const long long mod = 998244353;
  long long fact[N];
  map<long long, long long> vis;
  long long ksm(long long a, long long b) {
    if (b == 0) return 1;
    if (b & 1)
      return a * ksm(a, b - 1) % mod;
    else {
      long long k = ksm(a, b / 2);
      return k * k % mod;
    }
  }
  long long C(long long m, long long n) {
    if (m == 0) return 1;
    return fact[n] * ksm(fact[m] * fact[n - m] % mod, mod - 2) % mod;
  }
  signed main() {
    fact[0] = 1;
    for (long long i = 1; i < N; i++) {
      fact[i] = fact[i - 1] * i % mod;
    }
    long long T;
    cin >> T;
    while (T--) {
      long long n;
      cin >> n;
      vis.clear();
      long long mx = 0;
      for (long long i = 1; i <= n; i++) {
        long long x;
        cin >> x;
        vis[x]++;
        mx = max(x, mx);
      }
      if (vis[mx] > 1)
        cout << fact[n] << endl;
      else {
        if (vis[mx - 1] == 0) {
          cout << 0 << endl;
        } else {
          long long all = n - vis[mx] - vis[mx - 1];
          long long res = fact[n];
          for (long long i = vis[mx - 1] + 1; i <= n; i++) {
            res = ((res - C(n - i, all) * fact[i - 1] % mod * fact[n - i] % mod) +
                   mod) %
                  mod;
          }
          cout << res << endl;
        }
      }
    }
  }
  