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 = 1000000007;
  const long long int qw = 998244353;
  long long int power(long long int base, long long int exp, long long int M) {
    base %= M;
    long long int res = 1;
    while (exp > 0) {
      if (exp & 1) res = (res * base) % M;
      base = (base * base) % M;
      exp = exp >> 1;
    }
    return res;
  }
  long long int fac(long long int n, long long int M) {
    long long int ans = 1;
    if (n == 0 || n == 1) {
      return 1;
    }
    for (long long int i = 2; i < n + 1; i++) {
      ans = (ans * i) % M;
    }
    return ans;
  }
  long long int inverse(long long int n, long long int M) {
    long long int fa = fac(n, qw);
    return power(fa, M - 2, M);
  }
  long long int ncr(long long int n, long long int r, long long int M) {
    long long int num = fac(n, qw);
    long long int p1 = inverse(r, qw);
    long long int p2 = inverse(n - r, qw);
    long long int den = (p1 * p2) % qw;
    long long int ans = (num * den) % qw;
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      long long int arr[n];
      long long int k, m;
      for (long long int i = 0; i < n; i++) {
        cin >> arr[i];
      }
      priority_queue<long long int> pq(arr, arr + n);
      m = pq.top();
      pq.pop();
      if (pq.top() == m) {
        cout << fac(n, qw) << "\n";
      } else {
        k = pq.top();
        if (k != m - 1) {
          cout << "0\n";
          continue;
        }
        long long int count = 0;
        while (pq.empty() == false && pq.top() == k) {
          count++;
          pq.pop();
        }
        long long int nfac = fac(n, qw);
        long long int ans = fac(count, qw);
        ans = (ans * fac(n - (count + 1), qw)) % qw;
        ans = (ans * ncr(n, count + 1, qw)) % qw;
        long long int f = (((nfac - ans) % qw) + qw) % qw;
        cout << f << "\n";
      }
    }
    return 0;
  }
  