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 long long N = 1e6 + 9;
  const long long inf = 0x3f3f3f3f;
  const long long mod = 998244353;
  long long a[N];
  long long fac[N], inv_fac[N];
  long long q_pow(long long a, long long b, long long mod) {
    long long ret = 1;
    for (long long x = a; b; b >>= 1, (x *= x) %= mod)
      if (b & 1) (ret *= x) %= mod;
    return ret;
  }
  long long inv(long long x) { return q_pow(x, mod - 2, mod); }
  void init() {
    fac[0] = inv_fac[0] = 1;
    for (int i = 1; i < N; i++) {
      fac[i] = (fac[i - 1] * 1ll * i) % mod;
      inv_fac[i] = inv(fac[i]);
    }
  }
  long long C(long long n, long long m) {
    return fac[n] * inv_fac[m] % mod * inv_fac[n - m] % mod;
  }
  void solve() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    sort(a + 1, a + 1 + n, greater<int>());
    long long ma1 = 1, ma2 = 0;
    for (int i = 2; i <= n; i++) {
      if (a[i] == a[1])
        ma1++;
      else if (a[i] + 1 == a[1])
        ma2++;
      else {
        if (ma2 == 0 && ma1 == 1) {
          cout << 0 << endl;
          return;
        }
        break;
      }
    }
    if (ma1 > 1) {
      cout << fac[n] << endl;
    } else {
      cout << fac[n] * inv_fac[ma2 + 1] % mod * ma2 % mod * fac[ma2] % mod
           << endl;
    }
  }
  signed main() {
    init();
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  