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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  using namespace std;
  const long long int mod = 998244353;
  long long int power(long long int x, long long int y) {
    long long int res = 1;
    x = x % mod;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % mod;
      y = y >> 1;
      x = (x * x) % mod;
    }
    return res;
  }
  long long int inv(long long int a) { return power(a, mod - 2); }
  long long int add(long long int a, long long int b) { return (a + b) % mod; }
  long long int sub(long long int a, long long int b) {
    return (a - b + mod) % mod;
  }
  long long int mul(long long int a, long long int b) { return (a * b) % mod; }
  long long int divi(long long int a, long long int b) {
    return (a * inv(b)) % mod;
  }
  long long fact[300000 + 1];
  void pre() {
    fact[0] = 1;
    for (int i = 1; i < 300000; i++) fact[i] = mul(fact[i - 1], i);
  }
  void solve() {
    int n;
    cin >> n;
    long long arr[n];
    map<long long, long long> m;
    long long ma = 0;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      m[arr[i]]++;
      ma = max(arr[i], ma);
    }
    if (m[ma] > 1)
      cout << fact[n] << "\n";
    else
      cout << sub(fact[n], divi(fact[n], m[ma] + m[ma - 1])) << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    pre();
    long long int num = 1;
    cin >> num;
    for (long long int i = 0; i < num; i++) {
      solve();
    }
  }
  