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;
  using ll = long long;
  using pll = pair<ll, ll>;
  using t4 = tuple<ll, ll, ll, ll>;
  int T;
  const int mod = 998244353;
  ll fact[220000];
  ll invfact[220000];
  ll modpow(ll a, ll k) {
    if (k == 0) return 1;
    if (k & 1)
      return modpow(a, k - 1) * a % mod;
    else {
      ll half = modpow(a, k / 2);
      return half * half % mod;
    }
  }
  ll nCk(ll n, ll k) {
    if (k == 0 || n == k) return 1;
    return fact[n] * invfact[k] % mod * invfact[n - k] % mod;
  }
  void f() {
    int N;
    cin >> N;
    vector<ll> v(N);
    for (int i = 0; i < N; i++) cin >> v[i];
    ll mx = *max_element(v.begin(), v.end());
    int cnt = 0;
    for (int i = 0; i < N; i++)
      if (mx == v[i]) cnt++;
    if (cnt >= 2) {
      cout << fact[N] << '\n';
    } else {
      int mx2 = 0, cnt2 = 0;
      for (int i = 0; i < N; i++) {
        if (v[i] < mx && v[i] > mx2) {
          mx2 = v[i];
          cnt2 = 1;
        } else if (mx2 == v[i])
          cnt2++;
      }
      if (mx2 != mx - 1) {
        cout << 0 << '\n';
        return;
      }
      ll ans = fact[N];
      ll tmp = fact[cnt2] * fact[N - cnt2 - 1] % mod;
      tmp = tmp * nCk(N, N - cnt2 - 1) % mod;
      ans = (ans - tmp + mod) % mod;
      cout << ans << '\n';
    }
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin >> T;
    fact[0] = 1;
    for (int i = 1; i <= 200000; i++) fact[i] = fact[i - 1] * i % mod;
    invfact[200000] = modpow(fact[200000], mod - 2);
    for (int i = 200000; i >= 1; i--) invfact[i - 1] = invfact[i] * i % mod;
    while (T--) f();
    return 0;
  }
  