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;
  bool sortbyfir(const pair<int, int> &a, const pair<int, int> &b) {
    if (a.first == b.first) return a.second < b.second;
    return a.first < b.first;
  }
  bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
    if (a.second == b.second) return a.first < b.first;
    return (a.second < b.second);
  }
  bool sr(const pair<int, int> &a, const pair<int, int> &b) {
    if (a.first == b.first)
      return a.second > b.second;
    else
      return a.first < b.first;
  }
  bool compmx(int a, int b) { return (a < b); }
  long long int gcd(long long int a, long long int b) {
    if (b == 0)
      return a;
    else
      return gcd(b, a % b);
  }
  long long int lcm(long long int a, long long int b) {
    return (a / gcd(a, b)) * b;
  }
  bool is_prime(long long int x) {
    if (x == 1) return false;
    if (x == 2 || x == 3) return true;
    for (long long int i = 2; i * i <= x; i++) {
      if (x % i == 0) return false;
    }
    return true;
  }
  vector<long long int> primes;
  void getPrimes() {
    vector<bool> p(1001, true);
    for (long long int i = 2; i * i <= 1000; i++) {
      if (p[i]) {
        for (long long int j = i * i; j <= 1000; j += i) {
          p[j] = false;
        }
      }
    }
    for (long long int i = 2; i <= 1000; i++) {
      if (p[i]) {
        primes.push_back(i);
      }
    }
  }
  long long int dx[] = {-1, 0, 1, 0};
  long long int dy[] = {0, -1, 0, 1};
  long long int const N = 2 * 100000 + 1;
  long long int power(long long int a, long long int b) {
    long long int res = 1ll;
    while (b > 0) {
      if (b % 2ll) res = (res * a) % 998244353;
      a = (a * a) % 998244353;
      b /= 2ll;
    }
    return res;
  }
  long long int fact[N], invfact[N];
  void factorial() {
    fact[0] = fact[1] = 1;
    for (long long int i = 2; i < N; i++) fact[i] = (i * fact[i - 1]) % 998244353;
    invfact[N - 1] = power(fact[N - 1], 998244353 - 2);
    for (long long int i = N - 2; i >= 0; i--)
      invfact[i] = (invfact[i + 1] * (i + 1)) % 998244353;
  }
  long long int nCr(long long int n, long long int r) {
    return (((fact[n] * invfact[r]) % 998244353) * invfact[n - r]) % 998244353;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int Q;
    cin >> Q;
    factorial();
    while (Q--) {
      long long int n;
      cin >> n;
      vector<long long int> a(n);
      for (long long int i = 0; i < n; i++) cin >> a[i];
      sort((a).begin(), (a).end());
      if (a[n - 1] - a[n - 2] == 0)
        cout << fact[n] % 998244353 << endl;
      else if (a[n - 1] - a[n - 2] == 1) {
        long long int c = 0;
        for (long long int i = 0; i < n; i++)
          if (a[i] == a[n - 2]) c++;
        long long int l = nCr(n, c + 1) % 998244353;
        l = (l * fact[c]) % 998244353;
        l = (l * fact[n - (c + 1)]) % 998244353;
        long long int ans = (fact[n] % 998244353 - l + 998244353) % 998244353;
        cout << ans << endl;
      } else
        cout << 0 << endl;
    }
    return 0;
  }
  