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 = 2e5 + 5;
  const long long MOD = 998244353;
  long long fact[N], invfact[N];
  long long pow(long long a, long long b, long long m) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans = (ans * a) % m;
      b /= 2;
      a = (a * a) % m;
    }
    return ans;
  }
  long long modinv(long long k) { return pow(k, MOD - 2, MOD); }
  void precompute() {
    fact[0] = fact[1] = 1;
    for (long long i = 2; i < N; i++) {
      fact[i] = fact[i - 1] * i;
      fact[i] %= MOD;
    }
    invfact[N - 1] = modinv(fact[N - 1]);
    for (long long i = N - 2; i >= 0; i--) {
      invfact[i] = invfact[i + 1] * (i + 1);
      invfact[i] %= MOD;
    }
  }
  long long nCr(long long x, long long y) {
    if (y > x) return 0;
    long long num = fact[x];
    num *= invfact[y];
    num %= MOD;
    num *= invfact[x - y];
    num %= MOD;
    return num;
  }
  long long n;
  long long a[N];
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    precompute();
    long long t;
    cin >> t;
    while (t--) {
      cin >> n;
      map<long long, long long> m;
      for (long long i = 1; i <= n; i++) {
        cin >> a[i];
        m[a[i]]++;
      }
      sort(a + 1, a + n + 1);
      long long mx = a[n];
      long long mx2 = a[n - 1];
      if (mx == mx2) {
        long long ans = fact[n];
        cout << ans << "\n";
      } else if (mx > mx2 + 1)
        cout << 0 << "\n";
      else {
        long long totalWays = fact[n];
        long long invalidWays = fact[n] * modinv(m[mx2] + 1);
        invalidWays %= MOD;
        long long ans = (totalWays - invalidWays + MOD) % MOD;
        cout << ans << "\n";
      }
    }
    return 0;
  }
  