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 int mod = 998244353;
  long long fac[200010], infact[200010];
  long long qmi(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b % 2) {
        res = res * a % mod;
      }
      b >>= 1;
      a = a * a % mod;
    }
    return res;
  }
  void init() {
    fac[0] = 1;
    infact[0] = 1;
    for (int i = 1; i <= 200000; i++) {
      fac[i] = i * fac[i - 1] % mod;
      infact[i] = qmi(fac[i], mod - 2) % mod;
    }
  }
  int main() {
    int t;
    cin >> t;
    init();
    while (t--) {
      int fir[2] = {0}, sec[2] = {0};
      int n;
      cin >> n;
      for (int i = 0; i < n; i++) {
        int num;
        cin >> num;
        if (sec[0] == num) {
          sec[1]++;
        } else if (fir[0] == num) {
          fir[1]++;
        } else if (fir[0] < num) {
          sec[0] = fir[0];
          sec[1] = fir[1];
          fir[0] = num;
          fir[1] = 1;
        } else if (num > sec[0]) {
          sec[0] = num;
          sec[1] = 1;
        }
      }
      if (fir[1] > 1) {
        cout << fac[n] << endl;
        continue;
      } else if (fir[0] - sec[0] >= 2) {
        cout << 0 << endl;
        continue;
      } else {
        long long ans;
        ans = (fac[n] * infact[n - sec[1] - 1] % mod * infact[sec[1] + 1] % mod +
               mod) %
              mod;
        int k =
            fac[sec[1]] * (sec[1]) % mod * fac[n - sec[1] - 1] % mod * ans % mod;
        cout << k % mod << endl;
      }
    }
    return 0;
  }
  