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;
  template <class T, class... U>
  void re(T &&a, U &&...b) {
    cin >> forward<T>(a);
    (long long int[]){(cin >> forward<U>(b), 0)...};
  }
  template <class T, class... U>
  void pr(T &&a, U &&...b) {
    cout << forward<T>(a);
    (long long int[]){(cout << " " << forward<U>(b), 0)...};
    cout << " ";
  }
  void _print(long long 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 << "]";
  }
  template <class T>
  void print(const T *const arr, long long int count) {
    for (long long int i = 0; i < count; ++i) {
      _print(arr[i]);
      cerr << " ";
    }
    cerr << '\n';
  }
  long long int gcd(long long int a, long long int b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long int expo(long long int a, long long int b, long long int mod) {
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  void extendgcd(long long int a, long long int b, long long int *v) {
    if (b == 0) {
      v[0] = 1;
      v[1] = 0;
      v[2] = a;
      return;
    }
    extendgcd(b, a % b, v);
    long long int x = v[1];
    v[1] = v[0] - v[1] * (a / b);
    v[0] = x;
    return;
  }
  void swap(long long int &x, long long int &y) {
    long long int temp = x;
    x = y;
    y = temp;
  }
  long long int mod_add(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long int mod_mul(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long int mod_sub(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long int inverse(long long int a, long long int p) {
    return expo(a, p - 2, p);
  }
  long long int mod_div(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (mod_mul(a, inverse(b, m), m) + m) % m;
  }
  long long int fact[200001];
  void f() {
    fact[0] = 1, fact[1] = 1;
    for (long long int i = 2; i < 200001; i++) {
      fact[i] = mod_mul(i, fact[i - 1], 998244353);
    }
  }
  long long int ncr(long long int n, long long int r, long long int mod) {
    if (r == 0) return 1;
    return (fact[n] * inverse(fact[r], mod) % mod * inverse(fact[n - r], mod) %
            mod) %
           mod;
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int T = 1;
    cin >> T;
    f();
    for (long long int i = 1; i < T + 1; i++) {
      long long int n;
      re(n);
      long long int a[n], b[n];
      map<long long int, long long int> m;
      for (long long int i = 0; i < n; i++) {
        re(a[i]);
        b[i] = a[i];
        m[a[i]]++;
      }
      sort(b, b + n);
      long long int mx = b[n - 1], mx1 = -1;
      for (long long int i = n - 1; i >= 0; i--) {
        if (b[i] != mx) {
          mx1 = b[i];
          break;
        }
      }
      long long int ans = fact[n];
      long long int k = m[mx1];
      long long int x = n - 1 - k;
      long long int z = 0;
      if (m[mx] > 1) {
        pr(fact[n] % 998244353);
        goto label;
      }
      if (mx1 != -1 && mx - mx1 > 1) {
        cout << 0;
        goto label;
      }
      for (long long int i = 0; i < x + 1; i++) {
        long long int y =
            ((ncr(x, i, 998244353) % 998244353) * (fact[n - i - 1] % 998244353)) %
            998244353;
        y = (y * fact[i]) % 998244353;
        z += y;
        z %= 998244353;
      }
      if (n >= 13) ans += 998244353;
      pr(mod_sub(ans, z, 998244353));
    label:;
      cout << '\n';
      cout << '\n';
    }
    return 0;
  }
  