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;
  long long int fact[200009];
  long long int M = 998244353;
  void fac() {
    fact[0] = 1;
    for (long long int i = 1; i < 200003; i++) {
      fact[i] = fact[i - 1] * (i);
      fact[i] %= M;
    }
  }
  long long int D, X, Y;
  void extendedEuclid(long long int A, long long int B) {
    if (B == 0) {
      D = A;
      X = 1;
      Y = 0;
    } else {
      extendedEuclid(B, A % B);
      int temp = X;
      X = Y;
      Y = temp - (A / B) * Y;
    }
  }
  long long int modInverse(long long int A) {
    extendedEuclid(A, M);
    return (X % M + M) % M;
  }
  void solve() {
    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());
    long long int cnt = 0;
    if ((a[n - 1] - a[n - 2]) >= 2) {
      cout << "0\n";
      return;
    }
    map<long long int, long long int> m;
    cnt = 1;
    set<long long int> ss;
    for (long long int i = 0; i < n; i++) {
      m[a[i]]++;
      ss.insert(a[i]);
    }
    cnt = fact[n];
    if (m[a[n - 1]] > 1) {
      cout << cnt << "\n";
      return;
    }
    long long int c;
    c = n - 1 - m[a[n - 2]];
    cnt = 0;
    for (long long int i = 0; i < c + 1; i++) {
      long long int pp = ((fact[i]) * (fact[n - i - 1]));
      long long int pp2 = (fact[c]);
      long long int pp3 = (fact[i] * fact[c - i]);
      pp2 %= M;
      pp3 %= M;
      pp %= M;
      pp3 = modInverse(pp3);
      pp2 *= pp3;
      pp2 %= M;
      cnt = (cnt + (pp * pp2) + M);
      ;
      cnt %= M;
    }
    cnt = (fact[n] - cnt + M);
    cnt %= M;
    cout << cnt << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    fac();
    long long int t = 1;
    cin >> t;
    for (long long int i = 1; i <= t; i++) {
      solve();
    }
  }
  