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;
  inline bool is_digit(int ch) { return ch >= '0' && ch <= '9'; }
  inline bool is_int(int ch) { return is_digit(ch) || ch == '-' || ch == '+'; }
  inline int next_int() {
    int ch, sign = 1;
    int val = 0;
    while (!is_int(ch = getchar()) && ch != (-1))
      ;
    if (is_digit(ch)) {
      val = (int)ch - '0';
    } else if (ch == '-') {
      sign = -1;
    }
    while (is_int(ch = getchar())) {
      val = val * 10 + ch - '0';
    }
    return sign * val;
  }
  inline void print_int(int n) {
    static char print_buffer[0x20];
    size_t front = 0, sign = 0;
    if (n < 0) {
      n = (~n) + 1;
      sign = !sign;
    }
    do {
      print_buffer[front++] = (n % 10) + '0';
      n /= 10;
    } while (n);
    if (sign) {
      print_buffer[front++] = '-';
    }
    while (front--) {
      putchar(print_buffer[front]);
    }
  }
  inline void print_int_ln(int n) {
    print_int(n);
    putchar('\n');
  }
  void print_ints(int n) { print_int(n); }
  template <typename... T>
  void print_ints(int first, T... values) {
    print_int(first);
    putchar(' ');
    print_ints(values...);
  }
  void print_ints_ln(int n) { print_int_ln(n); }
  template <typename... T>
  void print_ints_ln(int first, T... values) {
    print_int(first);
    putchar(' ');
    print_ints_ln(values...);
  }
  void print_int_array(int* arr, size_t length) {
    for (size_t i = 0; i != length; ++i) {
      print_int(arr[i]);
      putchar(i != length - 1 ? ' ' : '\n');
    }
  }
  inline void receive_int_array(int* arr, size_t length) {
    for (size_t i = 0; i != length; ++i) {
      arr[i] = next_int();
    }
  }
  inline size_t receive_line(char* str) {
    size_t length;
    int ch;
    for (length = 0; (ch = getchar()) != '\n' && ch != (-1); ++length) {
      str[length] = ch;
    }
    str[length] = 0;
    return length;
  }
  int n;
  int arr[200010];
  long long fac[200010] = {1};
  int main() {
    int cases = next_int();
    for (int i = 1; i < 200010; ++i) {
      fac[i] = (fac[i - 1] * i) % 998244353;
    }
    while (cases--) {
      n = next_int();
      receive_int_array(arr, n);
      int maxi = 0, sec = 0;
      for (int i = 0; i < n; ++i) {
        if (arr[i] > maxi) {
          sec = maxi;
          maxi = arr[i];
        } else if (arr[i] < maxi && arr[i] > sec) {
          sec = arr[i];
        }
      }
      int countsM = 0, countsS = 0;
      for (int i = 0; i < n; ++i) {
        countsM += arr[i] == maxi;
        countsS += arr[i] == sec;
      }
      if (countsM == 1 && maxi - sec > 1) {
        printf("0\n");
      } else {
        if (countsM > 1) {
          printf("%d\n", (int)fac[n]);
        } else {
          int rud = n - countsM - countsS;
          long long result = fac[n], acc = 1;
          for (int i = 0; i <= rud; ++i) {
            result = (result - fac[n - i - 1] * acc) % 998244353;
            acc = (acc * ((long long)rud - i)) % 998244353;
          }
          long long mod = 998244353;
          while (result < 0) {
            result += mod;
            mod *= 2;
          }
          result = result % 998244353;
          printf("%d\n", (int)result);
        }
      }
    }
    return 0;
  }
  