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 int MOD = 998244353;
  template <typename T>
  T add(T x) {
    return x;
  }
  template <typename T, typename... Ts>
  T add(T x, Ts... y) {
    T res = x + add(y...);
    if (res >= MOD) res -= MOD;
    return res;
  }
  template <typename T, typename... Ts>
  T sub(T x, Ts... y) {
    return add(x, MOD - add(y...));
  }
  template <typename T, typename... Ts>
  void udd(T& x, Ts... y) {
    x = add(x, y...);
  }
  template <typename T>
  T mul(T x) {
    return x;
  }
  template <typename T, typename... Ts>
  T mul(T x, Ts... y) {
    return (x * 1ll * mul(y...)) % MOD;
  }
  template <typename T, typename... Ts>
  void uul(T& x, Ts... y) {
    x = mul(x, y...);
  }
  int bin(long long a, long long deg) {
    int r = 1;
    while (deg) {
      if (deg & 1) uul(r, a);
      deg >>= 1;
      uul(a, a);
    }
    return r;
  }
  long long inv(long long x) {
    assert(x);
    return bin(x, MOD - 2);
  }
  vector<long long> fac(2 * 1e5 + 1);
  void precalc() {
    fac[0] = 1;
    for (int i = 1; i <= 2 * 1e5; ++i) {
      fac[i] = mul(fac[i - 1], i);
    }
  }
  long long solve(size_t n, vector<int>& a) {
    long long maxx = a[0];
    long long maxx_cnt = 1;
    long long second_max = -1;
    long long second_max_cnt = 0;
    for (int i = 1; i < n; ++i) {
      if (a[i] > maxx) {
        second_max = maxx;
        second_max_cnt = maxx_cnt;
        maxx = a[i];
        maxx_cnt = 1;
      } else if (a[i] == maxx) {
        ++maxx_cnt;
      } else if (second_max < a[i] && a[i] < maxx) {
        second_max = a[i];
        second_max_cnt = 1;
      } else if (a[i] == second_max) {
        ++second_max_cnt;
      }
    }
    if (second_max == -1) {
      return fac[n];
    }
    if (maxx_cnt > 1) {
      return fac[n];
    }
    if (second_max != maxx - 1) {
      return 0;
    }
    return mul(fac[n], inv(second_max_cnt + 1), second_max_cnt);
  }
  int getNext(vector<int>& a, int cur) {
    int n = a.size();
    int i = (cur + 1) % n;
    while (a[i] == 0) {
      i = (i + 1) % n;
      if (i == cur) {
        return cur;
      }
    }
    return i;
  }
  bool eval(vector<int>& a) {
    int i = 0;
    int next;
    int total_sum = 0;
    for (auto& val : a) {
      total_sum += val;
    }
    do {
      next = getNext(a, i);
      if (i == next) {
        if (total_sum == 1) {
          return true;
        }
        return false;
      }
      --a[i];
      --total_sum;
      i = next;
    } while (next != -1);
    return true;
  }
  long long getAns(vector<int>& initial_perm) {
    size_t n = initial_perm.size();
    vector<int> perm(n);
    for (int i = 0; i < n; ++i) {
      perm[i] = i;
    }
    long long ans = 0;
    vector<int> input(n);
    do {
      for (int i = 0; i < n; ++i) {
        input[i] = initial_perm[perm[i]];
      }
      if (eval(input)) {
        ans = (ans + 1) % MOD;
      }
    } while (next_permutation(perm.begin(), perm.end()));
    return ans;
  }
  int main() {
    precalc();
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t-- > 0) {
      int n;
      cin >> n;
      vector<int> a(n);
      for (auto& val : a) {
        cin >> val;
      }
      cout << solve(n, a) << endl;
    }
  }
  