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>
  #pragma GCC optimize("Ofast,unroll-loops")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  const long long mod = 998244353;
  const long long mod1 = 1e9 + 7;
  const long long MOD = 1e18 + 1e16;
  long long mod_mul(long long a, long long b) {
    a = a % mod;
    b = b % mod;
    return (((a * b) % mod) + mod) % mod;
  }
  long long inv(long long i) {
    if (i == 1) return 1;
    return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long pwr(long long a, long long b) {
    a %= mod;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return res;
  }
  long long findMinNumber(long long n) {
    long long count = 0, ans = 1;
    while (n % 2 == 0) {
      count++;
      n /= 2;
    }
    if (count % 2) ans *= 2;
    for (int i = 3; i <= sqrt(n); i += 2) {
      count = 0;
      while (n % i == 0) {
        count++;
        n /= i;
      }
      if (count % 2) ans *= i;
    }
    if (n > 2) ans *= n;
    return ans;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  vector<long long> fact(300005, 1);
  long long C(long long n, long long r) {
    long long ans = 1;
    ans = fact[n];
    ans = mod_mul(ans, pwr(fact[r], mod - 2));
    ans = mod_mul(ans, pwr(fact[n - r], mod - 2));
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t, i;
    cin >> t;
    for (i = 2; i < 300005; i++) {
      fact[i] = mod_mul(fact[i - 1], i);
    }
    while (t--) {
      long long n;
      cin >> n;
      long long a[n], i;
      for (i = 0; i < n; i++) {
        cin >> a[i];
      }
      sort(a, a + n);
      long long res = 0;
      if (a[n - 1] - a[n - 2] > 1) {
        cout << 0 << '\n';
      } else {
        long long ans = fact[n];
        long long cnt = 0;
        for (i = 0; i < n; i++)
          if (a[n - 1] - a[i] <= 1)
            break;
          else
            cnt++;
        if (a[n - 2] == a[n - 1])
          cout << ans % mod << '\n';
        else {
          long long ans1 = 0;
          for (i = 0; i < cnt + 1; i++) {
            ans1 += ((fact[n - 1 - i] * C(cnt, i)) % mod * fact[i]) % mod;
            ans1 %= mod;
          }
          cout << ((ans - ans1) % mod + mod) % mod << '\n';
        }
      }
    }
    return 0;
  }
  