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 int N = 1000001;
  long long int factorialNumInverse[N + 1];
  long long int naturalNumInverse[N + 1];
  long long int spf[N + 1];
  long long int fact[N + 1];
  vector<long long int> z_function(string s) {
    long long int n = (long long int)s.length();
    vector<long long int> z(n);
    for (long long int i = 1; i < n; ++i)
      while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    return z;
  }
  int maxSubArraySum(int a[], int size) {
    int max_so_far = INT_MIN, max_ending_here = 0;
    for (int i = 0; i < size; i++) {
      max_ending_here = max_ending_here + a[i];
      if (max_so_far < max_ending_here) max_so_far = max_ending_here;
      if (max_ending_here < 0) max_ending_here = 0;
    }
    return max_so_far;
  }
  vector<long long int> SieveOfEratosthenes(int n) {
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
    for (int p = 2; p * p <= n; p++) {
      if (prime[p] == true) {
        for (int i = p * p; i <= n; i += p) prime[i] = false;
      }
    }
    vector<long long int> v;
    for (int p = 2; p <= n; p++)
      if (prime[p]) v.push_back(p);
    return v;
  }
  void sieve() {
    spf[1] = 1;
    for (int i = 2; i < N; i++) spf[i] = i;
    for (int i = 4; i < N; i += 2) spf[i] = 2;
    for (int i = 3; i * i < N; i++) {
      if (spf[i] == i) {
        for (int j = i * i; j < N; j += i)
          if (spf[j] == j) spf[j] = i;
      }
    }
  }
  vector<int> getFactorization(int x) {
    vector<int> ret;
    while (x != 1) {
      ret.push_back(spf[x]);
      x = x / spf[x];
    }
    return ret;
  }
  void InverseofNumber() {
    naturalNumInverse[0] = naturalNumInverse[1] = 1;
    for (long long int i = 2; i <= N; i++)
      naturalNumInverse[i] = naturalNumInverse[998244353 % i] *
                             (998244353 - 998244353 / i) % 998244353;
  }
  void InverseofFactorial() {
    factorialNumInverse[0] = factorialNumInverse[1] = 1;
    for (long long int i = 2; i <= N; i++)
      factorialNumInverse[i] =
          (naturalNumInverse[i] * factorialNumInverse[i - 1]) % 998244353;
  }
  void factorial() {
    fact[0] = 1;
    for (long long int i = 1; i <= N; i++) {
      fact[i] = (fact[i - 1] * i) % 998244353;
    }
  }
  long long int power(long long int a, long long int b) {
    if (b == 0) return 1;
    long long int t = power(a, b / 2) % 998244353;
    long long int x = ((t % 998244353) * (t % 998244353)) % 998244353;
    if (b % 2) x = ((x % 998244353) * (a % 998244353)) % 998244353;
    return x;
  }
  long long int inverse(long long int x) { return power(x, 998244353 - 2); }
  long long int nCr(long long int n, long long int r) {
    long long int ans = ((fact[n] * factorialNumInverse[r]) % 998244353 *
                         factorialNumInverse[n - r]) %
                        998244353;
    return ans;
  }
  long long int nPr(long long int n, long long int r) {
    return (fact[n] * factorialNumInverse[n - r]) % 998244353;
  }
  long long int getparent(long long int a, vector<long long int> &parent) {
    return parent[a] = (parent[a] == a ? a : getparent(parent[a], parent));
  }
  void getunion(long long int a, long long int b, vector<long long int> &parent,
                vector<long long int> &ranking) {
    a = getparent(a, parent);
    b = getparent(b, parent);
    if (ranking[b] == ranking[a]) {
      parent[b] = a;
      ranking[a]++;
    } else if (ranking[b] > ranking[a]) {
      parent[a] = b;
    } else {
      parent[b] = a;
    }
  }
  long long int bintodec(string s) {
    long long int sum = 0;
    reverse(s.begin(), s.end());
    for (long long int i = 0; i < s.length(); i++) {
      sum += (s[i] == '1' ? pow(2, i) : 0);
    }
    return sum;
  }
  string dectobin(long long int n) {
    string s;
    while (n > 0) {
      if ((n & 1))
        s.push_back('1');
      else
        s.push_back('0');
      n = (n >> 1);
    }
    reverse(s.begin(), s.end());
    return s;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    InverseofNumber();
    InverseofFactorial();
    factorial();
    sieve();
    long long int t, n, i, x, y, m, p, q, sum, ans, cnt, mx, mn, j, k;
    cin >> t;
    while (t--) {
      sum = 0;
      cnt = 0;
      ans = 0;
      mx = INT_MIN;
      mn = 1e18;
      cin >> n;
      vector<long long int> a(n);
      for (i = 0; i < n; i++) cin >> a[i];
      sort(a.begin(), a.end());
      x = a[n - 1] - a[n - 2];
      if (x >= 2) {
        cout << 0 << "\n";
        continue;
      }
      if (x == 0) {
        cout << fact[n] << "\n";
        continue;
      }
      for (i = 0; i < n; i++) {
        if (a[i] == a[n - 2]) sum++;
      }
      ans = fact[n];
      for (i = 1; i <= n - sum; i++) {
        ans = (ans -
               (((fact[n - i] * nCr(n - sum - 1, i - 1)) % 998244353) *
                fact[i - 1]) %
                   998244353 +
               998244353) %
              998244353;
      }
      cout << ans << "\n";
    }
  }
  