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;
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  const long long MOD = 998244353;
  long long facts[200000 + 5];
  long long binpow(long long a, long long n) {
    long long res = 1;
    while (n) {
      if (n & 1) {
        res *= a;
        res %= MOD;
      }
      a *= a;
      a %= MOD;
      n >>= 1;
    }
    return res;
  }
  long double bin_search(long double num) {
    long double l = 0;
    long double r = num;
    while (r - l >= 1e-7) {
      long double m = (r + l) / 2;
      if (m * m > num) {
        r = m;
      } else {
        l = m;
      }
    }
    return l;
  }
  long long fact(long long n) { return (facts[n] + MOD) % MOD; }
  long long C(long long n, long long m) {
    long long top = fact(n);
    return (((((top * binpow(fact(m), MOD - 2)) + MOD) % MOD) *
                 binpow(fact(n - m), MOD - 2) +
             MOD) %
            MOD);
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> inp(n);
    long long cntMax = 0;
    long long _max = INT_MIN;
    set<long long> elems;
    map<long long, long long> cnts;
    for (long long i = 0; i < n; i++) {
      cin >> inp[i];
      elems.insert(inp[i]);
      if (inp[i] > _max) {
        cntMax = 0;
        _max = inp[i];
      }
      if (inp[i] == _max) {
        cntMax++;
      }
      cnts[inp[i]]++;
    }
    if (cntMax > 1) {
      cout << fact(n) << '\n';
      return;
    }
    elems.erase(_max);
    long long beforeMax = *elems.rbegin();
    if (_max - beforeMax > 1) {
      cout << 0 << '\n';
      return;
    }
    long long ans = fact(n);
    long long m = n - cnts[beforeMax] - 1;
    for (long long i = 0; i <= m; i++) {
      long long comb = C(m, i);
      long long firstFact = fact(i);
      long long secondFact = fact(n - i - 1);
      assert(comb > 0);
      assert(firstFact > 0);
      assert(secondFact > 0);
      long long tmp = (comb * firstFact) + MOD;
      tmp %= MOD;
      tmp *= secondFact;
      tmp %= MOD;
      ans -= tmp;
      ans += MOD;
      ans %= MOD;
      assert(ans > 0);
    }
    cout << ans << '\n';
  }
  int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    facts[0] = 1;
    for (long long i = 1; i < 200000 + 5; i++) {
      facts[i] = (facts[i - 1] * i) % MOD;
    }
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
    cerr << "\nTime: " << (double)clock() / CLOCKS_PER_SEC << endl;
    return 0;
  }
  