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 inf = 1e18;
  const int mod = 998244353;
  int add(int a, int b) {
    if (a + b >= mod) return a + b - mod;
    return a + b;
  }
  int mul(int a, int b) { return (a * 1ll * b) % mod; }
  int sub(int a, int b) {
    if (a - b >= 0)
      return a - b;
    else
      return a - b + mod;
  }
  int bin_pow(int a, int n) {
    if (n == 0) return 1;
    if (n % 2 == 0) {
      int b = bin_pow(a, n / 2);
      return mul(b, b);
    } else
      return mul(a, bin_pow(a, n - 1));
  }
  const int N = 2e5 + 1;
  int f[N], rf[N];
  int C(int k, int n) { return mul(f[n], mul(rf[k], rf[n - k])); }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    f[0] = f[1] = rf[0] = rf[1] = 1;
    for (int i = 2; i < N; i++) {
      f[i] = mul(f[i - 1], i);
      rf[i] = mul(rf[i - 1], bin_pow(i, mod - 2));
    }
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      map<int, int> cnt;
      for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        cnt[a]++;
      }
      if (cnt.rbegin()->second >= 2) {
        cout << f[n] << '\n';
      } else {
        auto it = (--(--(cnt.end())));
        if (it->first + 1 != cnt.rbegin()->first) {
          cout << 0 << '\n';
        } else {
          int c = it->second;
          int ans = mul(n - 1, f[n - 1]);
          for (int i = c; i < n - 1; i++) {
            ans = sub(ans, mul(C(c, i), mul(f[c], f[n - 1 - c])));
          }
          cout << ans << '\n';
        }
      }
    }
    return 0;
  }
  