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;
  long long MOD = 998244353;
  long long power(long long a, long long b) {
    long long result = 1;
    while (b > 0ll) {
      if (b % 2ll == 1ll) {
        result *= a;
      }
      a *= a;
      b /= 2ll;
    }
    return result;
  }
  long long countSetBits(long long x) {
    long long Count = 0;
    while (x > 0) {
      if (x & 1) Count++;
      x = x >> 1;
    }
    return Count;
  }
  bool isPerfectSquare(long long n) {
    long long sr = sqrt(n);
    if (sr * sr == n)
      return true;
    else
      return false;
  }
  long long mod(long long x, long long M) { return ((x % M + M) % M); }
  long long mul(long long a, long long b, long long M) {
    return mod(mod(a, M) * mod(b, M), M);
  }
  long long powerM(long long a, long long b, long long M) {
    long long res = 1ll;
    while (b) {
      if (b % 2ll == 1ll) {
        res = mul(a, res, M);
      }
      a = mul(a, a, M);
      b /= 2ll;
    }
    return res;
  }
  int log22(long long x) { return 64 - __builtin_clzll(x) - 1; }
  vector<int> SieveOfEratosthenes(int n) {
    vector<int> mem;
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
    for (long long p = 2; p * p <= n; p++) {
      if (prime[p] == true) {
        for (long long i = p * p; i <= n; i += p) prime[i] = false;
      }
    }
    for (long long i = 2; i < n; i++)
      if (prime[i]) mem.push_back(i);
    return mem;
  }
  long long modInverse(long long n, long long p) { return powerM(n, p - 2, p); }
  long long nCrMod(long long n, int r, long long p) {
    if (n < r) return 0;
    if (r == 0) return 1;
    long long fac[n + 1];
    fac[0] = 1;
    for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
           p;
  }
  long long phac(long long n) {
    long long res = 1;
    for (long long i = 1; i < n + 1; i++) (res *= i) %= MOD;
    return res;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<int> v(n + 1);
    int maxm = 0;
    int maxc = 0;
    for (long long i = 1; i < n + 1; i++) {
      cin >> v[i];
      maxm = max(maxm, v[i]);
    }
    for (long long i = 1; i < n + 1; i++) {
      if (v[i] == maxm) maxc++;
    }
    int secmax = 0;
    int secmaxc = 0;
    for (long long i = 1; i < n + 1; i++) {
      if (v[i] == maxm) continue;
      secmax = max(secmax, v[i]);
    }
    for (long long i = 1; i < n + 1; i++) {
      if (secmax == v[i]) secmaxc++;
    }
    int ans = phac(n);
    if (maxc > 1) {
      cout << ans << '\n';
      return;
    }
    if (maxm > secmax + 1 && secmax != 0) {
      cout << 0 << '\n';
      return;
    }
    long long res = nCrMod(n, secmaxc + 1, MOD);
    (res *= phac(secmaxc)) %= MOD;
    (res *= phac(n - secmaxc - 1)) %= MOD;
    (ans -= res) %= MOD;
    if (ans <= 0) ans += MOD;
    cout << ans % MOD << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  