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 long long N = 2e5 + 10;
  const long long mod = 998244353;
  long long a[N];
  long long rjc[N];
  long long jc[N];
  long long inverse(long long n) {
    long long ans = 1, p = 998244351;
    while (p) {
      if (p & 1) ans = (ans * n) % 998244353;
      n = (n * n) % 998244353;
      p >>= 1;
    }
    return ans;
  }
  void init() {
    long long res = 1;
    rjc[0] = jc[0] = 1;
    for (long long i = 1; i <= 2e5; i++) {
      jc[i] = (jc[i - 1] * i) % mod;
      rjc[i] = (rjc[i - 1] * inverse(i)) % mod;
    }
  }
  long long cal(long long m, long long n) {
    return ((jc[n] * rjc[n - m]) % mod) % mod;
  }
  signed main() {
    init();
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      for (long long i = 0; i < n; i++) {
        cin >> a[i];
      }
      long long ans = 0, cnt1 = 0, cnt2 = 0;
      sort(a, a + n, greater<long long>());
      long long maxx = a[0];
      for (long long i = 0; i < n; i++) {
        if (a[i] == maxx) cnt1++;
        if (a[i] == maxx - 1) cnt2++;
        if (a[i] < maxx - 1) break;
      }
      if (cnt1 >= 2)
        cout << jc[n] << '\n';
      else if (cnt1 < n && cnt2 == 0)
        cout << 0 << '\n';
      else {
        long long ans = jc[n];
        ans = (ans - jc[n - 1] + mod) % mod;
        long long cnt = n - cnt1 - cnt2;
        for (long long i = 1; i <= cnt; i++) {
          long long temp = cal(i, cnt) % mod;
          temp = (temp * jc[n - i - 1]) % mod;
          ans -= temp;
          ans = (ans + mod) % mod;
        }
        cout << ans << '\n';
      }
    }
    return 0;
  }
  