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 = 200000 + 9;
  const int P = 998244353;
  long long fact[N], Inv[N], inv_fact[N];
  void find_inv() {
    Inv[0] = 0;
    Inv[1] = 1;
    for (int i = 2; i < N; ++i) Inv[i] = (P - P / i) * Inv[P % i] % P;
    fact[0] = inv_fact[0] = 1;
    for (int i = 1; i < N; ++i) {
      fact[i] = fact[i - 1] * i % P;
      inv_fact[i] = inv_fact[i - 1] * Inv[i] % P;
    }
  }
  long long C(int n, int m) {
    return fact[n] * inv_fact[m] % P * inv_fact[n - m] % P;
  }
  void solve() {
    int n;
    scanf("%d", &n);
    map<int, int> mp;
    for (int i = 1; i <= n; ++i) {
      int a;
      scanf("%d", &a);
      ++mp[a];
    }
    if ((int)mp.size() == 1) {
      printf("%lld\n", fact[n]);
    } else if (mp.rbegin()->second >= 2) {
      printf("%lld\n", fact[n]);
    } else if (mp.rbegin()->first - (++mp.rbegin())->first == 1) {
      int x = (++mp.rbegin())->second;
      printf("%lld\n",
             (fact[x + 1] - fact[x] + P) * C(n, x + 1) % P * fact[n - x - 1] % P);
    } else {
      printf("0\n");
    }
  }
  int main() {
    find_inv();
    int T;
    scanf("%d", &T);
    while (T--) solve();
    return 0;
  }
  