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 double delta = 0.0000000001;
  const double pi = 3.1415926535;
  long long MOD = 998244353;
  void _print(long long 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 << "]";
  }
  void __print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void __print(T t, V... v) {
    _print(t);
    if (sizeof...(v)) cerr << ", ";
    __print(v...);
  }
  long long expo(long long a, long long b, long long mod) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) {
    long long x = gcd(a, b);
    long long ans = (a * b) / x;
    return ans;
  }
  long long extendedGCD(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long x1, y1;
    long long d = extendedGCD(b, a % b, x1, y1);
    x = y1;
    y = x1 - (a / b) * y1;
    return d;
  }
  void swap(long long &x, long long &y) {
    long long temp = x;
    x = y;
    y = temp;
  }
  long long modInverse(long long n, long long t) { return expo(n, t - 2, t); }
  void factors(long long n) {
    vector<long long> factor;
    for (long long i = 1; i * i <= n; i++)
      if (n % i == 0) {
        if (n / i == i)
          factor.push_back(i);
        else {
          factor.push_back(i);
          factor.push_back(n / i);
        }
      }
  }
  long long mod_sub(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  unsigned long long nCrModPFermat(long long n, long long r) {
    if (n < r) return 0;
    if (r == 0) return 1;
    vector<unsigned long long> fac(n + 1);
    fac[0] = 1;
    for (long long i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % MOD;
    return (fac[n] * modInverse(fac[r], MOD) % MOD * modInverse(fac[n - r], MOD) %
            MOD) %
           MOD;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> a(n);
    map<long long, long long> m;
    long long l = -1;
    long long tot = 1;
    for (long long i = (0); i <= (n - 1); i++) {
      cin >> a[i];
      m[a[i]]++;
      l = max(l, a[i]);
      tot = (tot * (i + 1)) % MOD;
    }
    if (n == 1) {
      if (a[0] == 1) {
        cout << 1 << "\n";
        return;
      } else {
        cout << 0 << "\n";
        return;
      }
    }
    if (m[l] > 1) {
      cout << (tot % MOD) << "\n";
      return;
    }
    tot = (tot * m[l - 1]) % MOD;
    tot = (tot * (modInverse(m[l - 1] + 1, MOD)) % MOD) % MOD;
    if (tot < 0) tot += MOD;
    cout << tot << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  