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 orz = 998244353LL;
  int a[200005];
  long long fac[200005] = {1, 1}, inv[200005] = {1, 1}, f[200005] = {1, 1};
  void genfact() {
    for (int i = 2; i < 200005; i++) {
      fac[i] = fac[i - 1] * i % orz;
      f[i] = (orz - orz / i) * f[orz % i] % orz;
      inv[i] = inv[i - 1] * f[i] % orz;
    }
  }
  long long C(long long n, long long m) {
    if (n < m) return 0;
    return fac[n] * inv[m] % orz * inv[n - m] % orz;
  }
  int main() {
    int T, i, n, j;
    long long ans, cnt, sub;
    scanf("%d", &T);
    genfact();
    while (T--) {
      scanf("%d", &n);
      ans = fac[n];
      for (i = 0; i < n; i++) {
        scanf("%d", &a[i]);
      }
      sort(a, a + n);
      for (i = n - 1, cnt = 0; i >= 0; i--) {
        if (a[i] == a[n - 1]) cnt++;
      }
      if (cnt > 1) {
        printf("%lld\n", ans);
        continue;
      }
      for (i = 0, cnt = 0; i < n; i++) {
        if (a[i] == a[n - 1] - 1) cnt++;
      }
      sub = ((C(n, cnt + 1) * fac[cnt] % orz) * fac[n - cnt - 1]) % orz;
      ans = (ans + orz - sub) % orz;
      printf("%lld\n", ans);
    }
    return 0;
  }
  