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;
  template <typename Head>
  inline void dout(Head in) {
    cerr << in << '\n';
  }
  long long bp(long long n, long long m, long long mod) {
    long long res = 1;
    while (m) {
      m & 1 ? res = res * n % mod : res;
      n = n * n % mod;
      m >>= 1LL;
    }
    return res;
  }
  int32_t main() {
    cin.tie(nullptr)->ios::sync_with_stdio(false);
    long long constexpr mod = (long long)998244353;
    long long constexpr maxn = (long long)2e5 + 1;
    vector<long long> fac(maxn);
    fac[0] = fac[1] = 1LL;
    for (long long i = 2; i < maxn; i++) {
      fac[i] = fac[i - 1] * (long long)i % mod;
    }
    vector<long long> inv(maxn);
    inv[maxn - 1] = bp(fac[maxn - 1], mod - 2, mod);
    for (long long i = maxn - 2; ~i; i--) {
      inv[i] = inv[i + 1] * (long long)(i + 1) % mod;
    }
    function<long long(long long, long long)> A = [&](long long l, long long r) {
      return fac[r] * inv[l] % mod;
    };
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<long long> a(n);
      map<long long, long long> cnt;
      for (long long &e : a) {
        cin >> e;
        cnt[e]++;
      }
      sort(a.begin(), a.end());
      if (cnt[a.back()] > 1) {
        cout << fac[n] << "\n";
        continue;
      }
      if (a.back() - a[(long long)a.size() - 2] > 1) {
        cout << "0\n";
        continue;
      }
      long long ans = fac[n];
      ans = (ans - fac[cnt[a[(long long)a.size() - 2]]] *
                       A(cnt[a[(long long)a.size() - 2]] + 1, n)) %
            mod;
      cout << (ans + mod) % mod << "\n";
    }
    return 0;
  }
  