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;
  vector<long long> fact(2000010);
  vector<long long> inv(2000010);
  const long long mod = 998244353;
  long long modpow(long long x, long long y) {
    return !y ? 1 : ((y % 2 ? x : 1) * modpow((x * x) % mod, y / 2)) % mod;
  }
  long long choose(long long x, long long y) {
    if (y > x) return 0;
    return (fact[x] * inv[y] % mod) * inv[x - y] % mod;
  }
  long long permute(long long x, long long y) {
    if (y > x) return 0;
    return (fact[x] * inv[x - y]) % mod;
  }
  void precalc() {
    fact[0] = 1;
    inv[0] = 1;
    for (long long i = 1; i <= 2000000; i++) {
      fact[i] = fact[i - 1] * i % mod;
    }
    inv[2000000] = modpow(fact[2000000], mod - 2);
    for (long long i = 1999999; i > 0; i--) {
      inv[i] = inv[i + 1] * (i + 1) % mod;
    }
  }
  signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    precalc();
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<long long> arr(n);
      map<long long, long long> freq;
      set<long long> s;
      for (long long i = 0; i < n; i++) {
        cin >> arr[i];
        s.insert(arr[i]);
        freq[arr[i]]++;
      }
      sort(arr.begin(), arr.end());
      if (arr[n - 2] == arr[n - 1]) {
        cout << fact[n] << "\n";
      } else if (arr[n - 2] + 1 == arr[n - 1]) {
        long long total = fact[n];
        long long rem = n - freq[arr[n - 1]] - freq[arr[n - 2]];
        long long val = fact[rem];
        long long cur = 0;
        for (long long i = 1; i <= n; i++) {
          cur += permute(i - 1, freq[arr[n - 2]]);
          cur %= mod;
        }
        total -= val * cur % mod;
        total += mod;
        total %= mod;
        cout << total << "\n";
      } else {
        cout << "0\n";
      }
    }
    return 0;
  }
  