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;
  using ll = long long;
  const ll mod = 998244353;
  inline namespace {
  template <typename B, typename E>
  inline B slow_pow(const B &b, const E &e) {
    assert(e >= 0);
    B res = 1;
    for (int q = 0; q < e; q++) res *= b;
    return res;
  }
  template <typename B, typename E>
  inline B bin_pow(const B &b, const E &e) {
    assert(e >= 0);
    B res = 1;
    B a = b;
    for (E q = e; q; q >>= 1) {
      if (q & 1) res *= a;
      a *= a;
    }
    return res;
  }
  template <typename B, typename E>
  inline B mod_pow(const B &b, const E &e) {
    assert(e >= 0);
    B res = 1;
    B a = b;
    for (E q = e; q; q >>= 1) {
      if (q & 1) res = (res * a) % mod;
      a = (a * a) % mod;
    }
    return res;
  }
  }  // namespace
  ll fact[200001];
  void pre() {
    fact[0] = 1;
    for (ll x = 1; x <= 200000; x++) fact[x] = (fact[x - 1] * x) % mod;
  }
  ll mod_inv(const ll &x) { return mod_pow(x, mod - 2); }
  ll choose(const int &n, const int &k) {
    return (((fact[n] * mod_inv(fact[k])) % mod) * mod_inv(fact[n - k])) % mod;
  }
  void solve() {
    int n;
    cin >> n;
    vector<ll> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    sort(a.begin(), a.end());
    ll ans;
    if (a[n - 1] == a[n - 2])
      ans = fact[n];
    else if (a[n - 1] == a[n - 2] + 1) {
      int cnt = 0;
      ans = 0;
      for (int i = 0; i < n - 1; i++) {
        if (a[i] + 1 == a[n - 1]) cnt++;
      }
      ans = (((choose(n, cnt + 1) * (((fact[cnt + 1] - fact[cnt]) + mod) % mod)) %
              mod) *
             fact[n - cnt - 1]) %
            mod;
    } else
      ans = 0;
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL);
    pre();
    int T = 1;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  