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 C1 = 1e9 + 7;
  const long long C2 = 998244353;
  void yes() { cout << "YES\n"; }
  void no() { cout << "NO\n"; }
  long long binpow(long long a, long long b, long long m = C1) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  vector<long long> fac(200001);
  vector<long long> invfac(200001);
  void solve() {
    long long n;
    cin >> n;
    vector<long long> v(n);
    long long cnt{};
    for (long long i = 0; i < n; i++) {
      cin >> v[i];
    }
    long long m = *max_element(v.begin(), v.end());
    map<long long, long long> freq;
    for (long long i = 0; i < n; i++) {
      if (m == v[i]) cnt++;
      freq[v[i]]++;
    }
    if (cnt == 1) {
      if (freq[m - 1] != 0) {
        long long x = freq[m - 1];
        long long ans = fac[n];
        ans -= ((fac[n] * binpow(x + 1, C2 - 2, C2)) % C2);
        ans += C2;
        ans %= C2;
        cout << ans << '\n';
      } else {
        cout << 0 << '\n';
      }
    } else {
      long long ans = fac[n];
      cout << ans << '\n';
    }
  }
  signed main() {
    long long t = 1;
    cin >> t;
    fac[0] = 1;
    fac[1] = 1;
    for (long long i = 2; i <= 2e5; i++) {
      fac[i] = (fac[i - 1] * i) % C2;
    }
    invfac[200000] = binpow(fac[200000], C2 - 2, C2);
    for (long long i = 200000 - 1; i >= 1; i--) {
      invfac[i] = (invfac[i + 1] * (i + 1)) % C2;
    }
    while (t--) {
      solve();
    }
  }
  