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 pack(1)
  #pragma GCC target("avx,avx2")
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  using namespace std;
  long long int fact(long long int n) {
    if (n == 0) return 1;
    return n * fact(n - 1) % 998244353;
  }
  long long int gcdExtended(long long int a, long long int b, long long int *x,
                            long long int *y);
  long long int modInverse(long long int b, long long int m) {
    long long int x, y;
    long long int g = gcdExtended(b, m, &x, &y);
    if (g != 1) return -1;
    return (x % m + m) % m;
  }
  long long int modDivide(long long int a, long long int b, long long int m) {
    a = a % m;
    long long int inv = modInverse(b, m);
    return (inv * a) % m;
  }
  long long int gcdExtended(long long int a, long long int b, long long int *x,
                            long long int *y) {
    if (a == 0) {
      *x = 0, *y = 1;
      return b;
    }
    long long int x1, y1;
    long long int gcd = gcdExtended(b % a, a, &x1, &y1);
    *x = y1 - (b / a) * x1;
    *y = x1;
    return gcd;
  }
  int main() {
    long long int t;
    cin >> t;
    while (t--) {
      long long int n;
      cin >> n;
      long long int a[n];
      for (long long int i = 0; i < n; ++i) cin >> a[i];
      sort(a, a + n);
      long long int cmax = 0;
      long long int csecmax = 0;
      long long int x = a[n - 1];
      for (long long int i = n - 1; i >= 0; --i) {
        if (a[i] != a[n - 1]) {
          x = a[i];
          break;
        }
      }
      for (long long int i = 0; i < n; ++i) {
        if (a[i] == a[n - 1])
          ++cmax;
        else if (a[i] == x)
          ++csecmax;
      }
      if (cmax > 1) {
        cout << fact(n) << endl;
      } else if ((a[n - 1] - x) > 1) {
        cout << "0" << endl;
      } else {
        long long int ans = 1;
        for (long long int i = csecmax + 1; i <= csecmax + cmax; ++i) {
          ans = ans * i % 998244353;
        }
        long long int fin =
            ((ans % 998244353) - (1 % 998244353) + 998244353) % 998244353;
        long long int z = ((fact(n) % 998244353) * (fin % 998244353)) % 998244353;
        cout << modDivide(z, ans, 998244353) << endl;
      }
    }
    return 0;
  }
  