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 N = 2e5 + 5, M = 998244353;
  int fact[N], invfact[N];
  int power(int x, int p) {
    int ret = 1;
    while (p) {
      if (p & 1) ret = (1LL * ret * x) % M;
      x = (1LL * x * x) % M;
      p >>= 1;
    }
    return ret;
  }
  int inv(int x) { return power(x, M - 2); }
  void go() {
    fact[0] = 1;
    invfact[0] = inv(fact[0]);
    for (int i = 1; i < N; i++) {
      fact[i] = (1LL * fact[i - 1] * i) % M;
      invfact[i] = inv(fact[i]);
    }
  }
  int nCr(int n, int c) {
    if (c > n) return 0;
    return 1LL * fact[n] * invfact[n - c] % M * invfact[c] % M;
  }
  int main() {
    go();
    int t;
    scanf("%d", &t);
    while (t--) {
      int n;
      scanf("%d", &n);
      vector<int> arr(n);
      for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
      sort(arr.rbegin(), arr.rend());
      if (arr[0] == arr[1]) {
        printf("%d\n", fact[n]);
        continue;
      }
      if (arr[0] - arr[1] > 1) {
        printf("0\n");
        continue;
      }
      int cnt = 0;
      for (int i = 0; i < n; ++i) {
        if (arr[1] == arr[i]) ++cnt;
      }
      int res = 0;
      for (int i = 1; i < n; i++) {
        res = (res + fact[n - 1]) % M;
        res =
            (res -
             (1LL * nCr(n - cnt - 1, n - i) * fact[i - 1] % M * fact[n - i] % M) +
             M) %
            M;
      }
      printf("%d\n", res);
    }
    return 0;
  }
  