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 double PI = 3.14159265389793238462643383279502884197169;
  const long long MOD = 1e9 + 7;
  const long long MOD1 = 998244353;
  const long long MODo = 1337377;
  const int MAXN = 2e5 + 2;
  const int oo = 1e9;
  const long long oo15 = 1e15;
  const long long oo18 = 1e18;
  const long long oooo = 9223372036854775807;
  int a[200005];
  long long fact[200005], infact[200005];
  long long power(long long x, long long y) {
    if (y == 0) return 1;
    long long yy = power(x, y / 2);
    if (y % 2)
      return yy * yy % MOD1 * x % MOD1;
    else
      return yy * yy % MOD1;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    fact[0] = 1;
    for (long long i = 1; i <= 2e5 + 3; i++) fact[i] = i * fact[i - 1] % MOD1;
    infact[200003] = power(fact[200003], MOD1 - 2);
    for (long long i = 2e5 + 2; i >= 0; i--)
      infact[i] = (i + 1) * infact[i + 1] % MOD1;
    int testcase;
    cin >> testcase;
    while (testcase--) {
      int n;
      cin >> n;
      int mx1 = 0, mx2 = 0;
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
        if (a[i] > mx1) {
          mx2 = mx1;
          mx1 = a[i];
        } else
          mx2 = max(mx2, a[i]);
      }
      if (mx1 - mx2 >= 2) {
        cout << 0 << "\n";
      } else if (mx2 == mx1) {
        cout << fact[n] << "\n";
      } else {
        long long dem = 0;
        for (int i = 1; i <= n; i++)
          if (a[i] == mx2) dem++;
        cout << fact[n - dem - 1] * fact[dem] % MOD1 *
                    (fact[n] * infact[dem + 1] % MOD1 * infact[n - dem - 1] %
                     MOD1) %
                    MOD1 * dem % MOD1
             << "\n";
      }
    }
  }
  