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 + 5;
  long long F[N];
  const long long M = 998244353;
  long long amodb(long long a) { return a - (a / M * M); }
  long long bigmod(long long a, long long b) {
    if (b == 0) return 1;
    long long ans = bigmod(a, b / 2);
    ans = amodb(ans * ans);
    if (b % 2) {
      ans = amodb(ans * a);
    }
    return ans;
  }
  void precal() {
    F[0] = 1;
    for (long long i = 1; i < N; i++) {
      F[i] = ((F[i - 1] % M) * (i % M)) % M;
    }
  }
  int main() {
    long long t = 1;
    precal();
    scanf("%lld", &t);
    while (t--) {
      long long n, i, j, k = 0, mx = 0, smx, cnt = 0, l, r, ans = 0;
      scanf("%lld", &n);
      vector<long long> v(n);
      for (i = 0; i < n; i++) {
        scanf("%lld", &v[i]);
      }
      sort(v.rbegin(), v.rend());
      if (v[0] - v[1] > 1) {
        printf("0\n");
      } else if (v[0] - v[1] == 0) {
        printf("%lld\n", F[n]);
      } else {
        for (i = 1; i < n; i++) {
          if (v[i] == v[1]) {
            k++;
          }
        }
        long long baad = (F[n] * bigmod(k + 1, M - 2)) % M;
        ans = (F[n] - baad) % M;
        ans += M;
        ans %= M;
        printf("%lld\n", ans);
      }
    }
  }
  