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_add(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long mod_mul(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long mod_sub(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  vector<long long> to_binary(long long n) {
    int i = 7;
    vector<long long> ans(8);
    while (n >= 2) {
      ans[i] = n % 2;
      n /= 2;
      i--;
    }
    ans[i] = n;
    return ans;
  }
  map<long long, long long> primefactorisation(long long n) {
    map<long long, long long> mp;
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        mp[i] = 0;
        while (n % i == 0) {
          mp[i]++;
          n /= i;
        }
      }
    }
    if (n > 1) {
      mp[n] = 1;
    }
    return mp;
  }
  vector<bool> seive(long long n) {
    vector<bool> arr(n + 1, true);
    for (int i = 2; i * i <= n; i++) {
      if (arr[i]) {
        for (int j = i * i; j <= n; j += i) {
          arr[j] = false;
        }
      }
    }
    return arr;
  }
  long long powM(long long x, long long y) {
    long long ans = 1, r = 1;
    x %= (long long)(998244353);
    while (r > 0 && r <= y) {
      if (r & y) {
        ans *= x;
        ans %= (long long)(998244353);
      }
      r <<= 1;
      x *= x;
      x %= (long long)(998244353);
    }
    return ans;
  }
  bool isprime(long long x) {
    for (long long i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        return 0;
      }
    }
    return 1;
  }
  bool cmp(pair<long long, long long> p1, pair<long long, long long> p2) {
    return (p1.first > p2.first);
  }
  void ipgraph(vector<set<long long>>& edge, long long v, long long e) {
    for (int i = 0; i < e; i++) {
      long long a, b;
      cin >> a >> b;
      edge[a].insert(b);
      edge[b].insert(a);
    }
  }
  void bfs(vector<vector<long long>>& edge, long long start,
           vector<bool>& visited) {
    long long curr = start;
    queue<long long> qq;
    qq.push(curr);
    visited[curr] = 1;
    while (!qq.empty()) {
      curr = qq.front();
      for (long long x : edge[curr]) {
        if (!visited[x]) {
          qq.push(x);
          visited[x] = 1;
        }
      }
      qq.pop();
    }
  }
  vector<string> bor = {"YES", "NO"};
  vector<long long> fact(300001);
  void pre() {
    fact[0] = 1;
    for (long long i = 1; i < fact.size(); i++) {
      fact[i] = mod_mul(i, fact[i - 1], (long long)(998244353));
    }
  }
  unsigned long long power(unsigned long long x, int y, int p) {
    unsigned long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  unsigned long long modInverse(unsigned long long n, int p) {
    return power(n, p - 2, p);
  }
  unsigned long long ncr(unsigned long long n, int r, int p) {
    if (n < r) return 0;
    if (r == 0) return 1;
    return (fact[n] * modInverse(fact[r], p) % p * modInverse(fact[n - r], p) %
            p) %
           p;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> arr(n);
    for (int i = 0; i < n; i++) cin >> arr[i];
    sort(arr.begin(), arr.end(), greater<long long>());
    long long maxx = arr[0];
    if (n == 1) {
      cout << 1 << "\n";
      return;
    }
    if (arr[1] == arr[0]) {
      cout << fact[n] % (long long)(998244353) << "\n";
      return;
    }
    if (arr[0] - arr[1] > 1) {
      cout << 0 << "\n";
      return;
    }
    long long x = 0;
    for (int i = 1; i < n; i++) {
      if (arr[i] == arr[1])
        x++;
      else
        break;
    }
    long long nf = fact[n];
    long long k = fact[n - x - 1];
    long long np = 0;
    long long diff = 1;
    long long ans = fact[n];
    for (int i = x + 1; i <= n; i++) {
      long long temp = ncr(i - 1, x, (long long)(998244353));
      temp = mod_mul(temp, fact[x], (long long)(998244353));
      temp = mod_mul(temp, k, (long long)(998244353));
      ans = mod_sub(ans, temp, (long long)(998244353));
      diff++;
    }
    cout << ans % (long long)(998244353) << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    pre();
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  