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;
  struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  const int64_t M = 998244353;
  const long long MAX = 200001;
  int64_t power(int64_t a, int64_t b) {
    b = (b < 0) ? (M + b - 1) : b;
    int64_t res = 1LL;
    while (b) {
      if (b % 2) res = res * a % M;
      a = a * a % M;
      b /= 2;
    }
    return res;
  }
  vector<long long> fac;
  void prepareFactorial() {
    fac.resize(MAX);
    fac[0] = fac[1] = 1;
    for (long long i = 2; i < MAX; i++) fac[i] = (fac[i - 1] * i) % M;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> jury(n);
    unordered_map<long long, long long, custom_hash> frequency;
    pair<long long, long long> max_pair;
    for (long long &task : jury) {
      cin >> task;
      frequency[task]++;
      if (task > max_pair.first) {
        swap(max_pair.first, max_pair.second);
        max_pair.first = task;
      } else if (task > max_pair.second) {
        max_pair.second = task;
      }
    }
    long long cntMx = frequency[max_pair.first];
    long long cntSecondMx = frequency[max_pair.second];
    long long ans = 0;
    if (cntMx > 1) {
      ans = fac[n];
    } else if (max_pair.first - max_pair.second == 1) {
      long long total = fac[n];
      long long invalid = (fac[n] * power(cntSecondMx + 1, -1)) % M;
      ans = (total - invalid + M) % M;
    }
    cout << ans << "\n";
  }
  int32_t main() {
    prepareFactorial();
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    long long t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  