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 int INF = 1000 * 1000 * 1000;
  const long long LINF = 1LL * INF * INF;
  const int MAX = 100010;
  const long double PI = acos(-1.);
  const double EPS = 1e-6;
  const long long MOD = 998244353;
  long long add(long long a, long long b) {
    long long f = a + b;
    if (f > MOD) f -= MOD;
    return f;
  }
  long long sub(long long a, long long b) {
    long long f = a - b;
    if (f < 0) f += MOD;
    return f;
  }
  long long mult(long long a, long long b) { return (a * b) % MOD; }
  long long binpow(long long a, long long n) {
    if (n == 0) return 1;
    if (n % 2) return mult(a, binpow(a, n - 1));
    long long b = binpow(a, n / 2);
    return mult(b, b);
  }
  long long inv(long long x) { return binpow(x, MOD - 2); }
  long long fast_inv(long long x, long long m) {
    long long u = 0, v = 1;
    while (x != 0) {
      long long t = m / x;
      m -= t * x;
      swap(x, m);
      u -= t * v;
      swap(u, v);
    }
    return u;
  }
  long long divide(long long a, long long b) { return mult(a, inv(b)); }
  long long dv(long long a, long long b) { return mult(a, fast_inv(b, MOD)); }
  long long fact[200500];
  long long C(int n, int k) {
    if (k < 0 || k > n) {
      return 0;
    }
    return divide(fact[n], mult(fact[k], fact[n - k]));
  }
  void prec() {
    fact[0] = 1;
    for (int i = 1; i < 200500; ++i) {
      fact[i] = mult(fact[i - 1], i);
    }
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (auto& a : v) cin >> a;
    ;
    map<int, int> m;
    for (int i = 0; i < n; ++i) {
      m[v[i]]++;
    }
    if (m.size() == 1) {
      cout << fact[n] << endl;
      return;
    }
    if (m.rbegin()->first - 1 > next(m.rbegin())->first &&
        m.rbegin()->second == 1) {
      cout << 0 << endl;
      return;
    }
    if (m.rbegin()->second > 1) {
      cout << fact[n] << endl;
      return;
    } else {
      int k = next(m.rbegin())->second;
      cout << sub(fact[n], mult(fact[k], mult(fact[n - k - 1], C(n, k + 1))))
           << endl;
    }
  }
  int main() {
    prec();
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  