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 = 2e5 + 1;
  long long fact[N];
  const int MOD = 998244353;
  long long power(long long x, long long y) {
    long long res = 1;
    x = x % MOD;
    while (y > 0) {
      if (y & 1) res = (res * x) % MOD;
      y = y >> 1;
      x = (x * x) % MOD;
    }
    return res;
  }
  long long modInverse(long long x) { return power(x, MOD - 2); }
  long long modAdd(long long a, long long b) {
    return (a % MOD + b % MOD + 2 * MOD) % MOD;
  }
  long long modMult(long long a, long long b) {
    return (a % MOD * b % MOD) % MOD;
  }
  long long modDiv(long long a, long long b) {
    return (a % MOD * modInverse(b)) % MOD;
  }
  void getFact() {
    fact[0] = 1;
    for (auto i = 1; i < N; i++) {
      fact[i] = (fact[i - 1] * i) % MOD;
    }
  }
  long long nCrModP(long long n, long long r) {
    if (r == 0) return 1;
    return ((fact[n] * modInverse(fact[r])) % MOD * modInverse(fact[n - r])) %
           MOD;
  }
  int main() {
    ios_base ::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int test;
    cin >> test;
    getFact();
    for (int t = 1; t <= test; t++) {
      int n;
      cin >> n;
      vector<long long> a(n);
      map<long long, int> m;
      for (int i = 0; i < n; i++) {
        cin >> a[i];
        m[a[i]]++;
      }
      sort(a.begin(), a.end());
      long long maxi = a[n - 1];
      int ind = n - 1;
      while (ind >= 0) {
        if (maxi == a[ind]) {
          ind--;
          continue;
        }
        if (maxi - a[ind] >= 2)
          maxi = -1;
        else
          maxi = a[ind];
        break;
      }
      if (maxi == -1 && m[a[n - 1]] == 1) {
        cout << "0\n";
        continue;
      }
      int cnt2 = m[maxi];
      int cnt1 = m[a[n - 1]];
      if (cnt1 > 1) {
        cout << fact[n] << "\n";
        continue;
      }
      int rem = n - cnt1 - cnt2;
      long long tot = fact[n];
      long long fs = fact[n - cnt1];
      fs = 0ll - fs;
      tot = modAdd(tot, fs);
      for (int i = 1; i <= rem; i++) {
        long long num = nCrModP(rem, i);
        long long rest = fact[n - 1 - i];
        long long no = fact[i];
        num = modMult(num, rest);
        no = modMult(no, num);
        no = 0ll - no;
        tot = modAdd(tot, no);
      }
      cout << tot << "\n";
    }
    return 0;
  }
  