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;
  ll KMOD = 998244353;
  ll mmul(ll a, ll b) { return a * b % KMOD; }
  ll extended_euclidean_algorithm(ll x, ll y, ll n_ax, ll n_mx, ll n_ay,
                                  ll n_my) {
    if (y == 1) return n_ay;
    ll rem = x % y;
    ll q = x / y;
    return extended_euclidean_algorithm(y, rem, n_ay, n_my, n_ax - n_ay * q,
                                        n_mx - n_my * q);
  }
  ll ModInverse(ll a, ll m) {
    return (extended_euclidean_algorithm(m, a % m, 0, 1, 1, 0) + m) % m;
  }
  ll mdiv(ll a, ll b) { return mmul(a, ModInverse(b, KMOD)); }
  vector<ll> factorial;
  void Preprocess() {
    factorial.resize(200001);
    factorial[0] = 1;
    factorial[1] = 1;
    for (int i = 2; i <= 200000; ++i) {
      factorial[i] = mmul(factorial[i - 1], i);
    }
  }
  int Solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    int max = *max_element((a).begin(), (a).end()), maxcnt = 0, max_m1_cnt = 0;
    for (int x : a) {
      if (x == max) {
        maxcnt++;
      } else if (x == max - 1) {
        max_m1_cnt++;
      }
    }
    if (maxcnt > 1) {
      return factorial[n];
    } else if (max_m1_cnt == 0) {
      return 0;
    } else {
      return mmul(factorial[n], mdiv(max_m1_cnt, max_m1_cnt + 1));
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    Preprocess();
    int t;
    cin >> t;
    while (t--) cout << Solve() << "\n";
  }
  