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 mod = 998244353;
  const int N = 1e6 + 4;
  void _print(long long t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  vector<long long> first(N, 1), INV(N, 1), minp(N, 0);
  vector<bool> prime(N, 0);
  long long gd(long long a, long long b) {
    if (b == 0) return a;
    return gd(b, a % b);
  }
  long long lcm(long long a, long long b) { return (a * b) / gd(a, b); }
  long long ex(long long x, long long n) {
    long long res = 1;
    x = x % mod;
    while (n) {
      if (n & 1) res = (res * x) % mod;
      x = (x * x) % mod;
      n = n >> 1;
    }
    return res % mod;
  }
  long long fermat(long long n) { return ex(n, mod - 2); }
  void fact() {
    for (int i = 2; i < N; i++) {
      first[i] = (first[i - 1] * i) % mod;
      INV[i] = fermat(first[i]);
    }
  }
  long long ncr(long long n, long long r) {
    if (r == 0 || n == r) return 1;
    if (r > n) return 0;
    return (first[n] * INV[r] % mod * INV[n - r] % mod) % mod;
  }
  void sieve() {
    prime[0] = 1, prime[1] = 1;
    for (int i = 2; i * i < N; i++) {
      if (prime[i] == 0) {
        for (int j = i * i; j < N; j += i) prime[j] = 1;
      }
    }
  }
  void minprime() {
    minp[0] = 0, minp[1] = 1;
    for (int i = 2; i < N; i++) {
      if (minp[i] == 0) {
        minp[i] = i;
        for (int j = i * i; j < N; j += i) {
          if (minp[j] == 0) minp[j] = i;
        }
      }
    }
  }
  void JAI_SHREE_RAM() {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    ;
    sort(a.begin(), a.end());
    map<long long, long long> m;
    for (int i = 0; i < n; i++) {
      m[a[i]]++;
    }
    if (abs(a[n - 1] - a[n - 2]) >= 2) {
      cout << 0 << "\n";
      return;
    }
    if (abs(a[n - 1] - a[n - 2]) == 0) {
      cout << first[n] << "\n";
      return;
    }
    long long invalid = first[n] * fermat(m[a[n - 2]] + 1) % mod;
    long long ans = (first[n] - invalid + mod) % mod;
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    fact();
    long long tttt = 1;
    cin >> tttt;
    while (tttt--) {
      JAI_SHREE_RAM();
    }
  }
  