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 fact[200001];
  long long power(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b % 2) {
        res = (res * a) % 998244353;
      }
      a = (a * a) % 998244353;
      b = b / 2;
    }
    return res;
  }
  long long solve(long long n, long long r) {
    if (r == 0) {
      return 1;
    }
    if (r > n) {
      return 1;
    }
    long long nfact = fact[n];
    long long nr = power(fact[n - r], 998244353 - 2);
    long long r1 = power(fact[r], 998244353 - 2);
    long long ans = nfact;
    ans = (ans * nr) % 998244353;
    ans = (ans * r1) % 998244353;
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    fact[0] = 1;
    fact[1] = 1;
    for (long long i = 2; i <= 200000; i++) {
      fact[i] = (i * fact[i - 1]) % 998244353;
    }
    int tc = 1;
    cin >> tc;
    while (tc--) {
      int n;
      cin >> n;
      int a[n];
      for (int i = 0; i < n; i++) {
        cin >> a[i];
      }
      sort(a, a + n);
      if (a[n - 1] - a[n - 2] > 1) {
        cout << "0"
             << "\n";
        continue;
      }
      if (a[n - 1] == a[n - 2]) {
        cout << fact[n] << "\n";
      } else {
        long long temp = 0;
        long long ct = 0;
        int el = a[n - 2];
        int i = n - 2;
        while (i >= 0 && a[i] == el) {
          i--;
          ct++;
        }
        for (int k = n; k > ct; k--) {
          long long elb = k - 1;
          long long ela = n - k;
          if (elb == n - 1) {
            temp += fact[n - 1];
          } else {
            long long nel = n - 1 - ct;
            long long plb = elb - ct;
            long long k1 = solve(nel, plb);
            long long toad = (k1 * fact[elb]) % 998244353;
            toad = (toad * fact[ela]) % 998244353;
            temp = (temp + toad) % 998244353;
          }
        }
        cout << (fact[n] - temp + 998244353) % 998244353 << "\n";
      }
    }
    return 0;
  }
  