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 vb = vector<bool>;
  using vvb = vector<vb>;
  using vi = vector<int>;
  using vvi = vector<vi>;
  using vl = vector<ll>;
  using vvl = vector<vl>;
  using vc = vector<char>;
  using vvc = vector<vc>;
  using vs = vector<string>;
  const ll mod = 998244353, inf = 1e18;
  void setIO() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
  }
  vl fact(2e5 + 5, 1);
  ll binPow(ll a, ll b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    ll ret = binPow(a, b / 2);
    if (b % 2 == 0) return (ret * ret) % mod;
    return ((ret * ret) % mod * a) % mod;
  }
  ll inv(ll a) { return (binPow(a, mod - 2) % mod + mod) % mod; }
  ll binom(ll a, ll b) {
    if (b < 0 or a < 0) return 0;
    return (((fact[a] * inv(fact[b])) % mod * inv(fact[a - b])) % mod + mod) %
           mod;
  }
  int main() {
    setIO();
    int t;
    cin >> t;
    for (int i = 1; i <= 2e5; i++) fact[i] = (i * fact[i - 1]) % mod;
    while (t--) {
      int n;
      cin >> n;
      vi a(n + 1);
      for (int i = 1; i <= n; i++) cin >> a[i];
      sort(a.begin() + 1, a.end());
      if (a[n] == a[n - 1]) {
        cout << fact[n] << '\n';
        continue;
      }
      if (a[n] > a[n - 1] + 1) {
        cout << 0 << '\n';
        continue;
      }
      ll cnt = 0;
      for (int i = 1; i <= n; i++) {
        if (a[i] == a[n - 1]) cnt++;
      }
      ll ans = ((fact[n] - ((fact[n - cnt - 1] * fact[cnt]) % mod *
                            binom(n, cnt + 1) % mod)) %
                    mod +
                mod) %
               mod;
      cout << ans << '\n';
    }
    return 0;
  }
  