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;
  void err(istream_iterator<string> it) {}
  template <typename T, typename... Args>
  void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << " ";
    err(++it, args...);
  }
  const long long MX = 2e5 + 105;
  const long long MOD = 998244353;
  inline void normal(long long &a) {
    a %= MOD;
    (a < 0) && (a += MOD);
  }
  inline long long modMul(long long a, long long b) {
    a %= MOD, b %= MOD;
    normal(a), normal(b);
    return (a * b) % MOD;
  }
  inline long long modAdd(long long a, long long b) {
    a %= MOD, b %= MOD;
    normal(a), normal(b);
    return (a + b) % MOD;
  }
  inline long long modSub(long long a, long long b) {
    a %= MOD, b %= MOD;
    normal(a), normal(b);
    a -= b;
    normal(a);
    return a;
  }
  inline long long modPow(long long b, long long p) {
    long long r = 1;
    while (p) {
      if (p & 1LL) r = modMul(r, b);
      b = modMul(b, b);
      p >>= 1LL;
    }
    return r;
  }
  inline long long modInverse(long long a) { return modPow(a, MOD - 2); }
  inline long long modDiv(long long a, long long b) {
    return modMul(a, modInverse(b));
  }
  long long fact[MX];
  long long ara[MX];
  void factorial() {
    fact[0] = 1;
    for (long long i = 1; i < MX; i++) fact[i] = modMul(fact[i - 1], i);
  }
  long long ncr(long long n, long long r) {
    if (n < r) return 0;
    long long ans = modMul(fact[r], fact[n - r]);
    ans = modDiv(fact[n], ans);
    return ans;
  }
  void solve() {
    long long n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> ara[i];
    }
    sort(ara, ara + n);
    if (ara[n - 2] == ara[n - 1]) {
      cout << fact[n] << endl;
      return;
    }
    if (ara[n - 1] - ara[n - 2] > 1) {
      cout << 0 << endl;
      return;
    }
    long long c = 1;
    for (int i = n - 3; i >= 0; i--) {
      if (ara[i] == ara[i + 1])
        c++;
      else
        break;
    }
    long long ans = ncr(n, c + 1);
    ans = modMul(ans, fact[n - c - 1]);
    ans = modMul(ans, fact[c]);
    ans = modSub(fact[n], ans);
    cout << ans << endl;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    factorial();
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  