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 MOD = 998244353;
  void chkmax(long long &a, long long b) {
    if (a < b) a = b;
  }
  void chkmin(long long &a, long long b) {
    if (a > b) a = b;
  }
  long long qpow(long long d, long long z) {
    long long res = 1;
    for (; z; z >>= 1) {
      if (z & 1) res = res * d % MOD;
      d = d * d % MOD;
    }
    return res;
  }
  int n;
  int d[200005];
  void solve() {
    scanf("%d", &n);
    int Max = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &d[i]);
      if (Max < d[i]) Max = d[i];
    }
    int cM0 = 0, cM1 = 0;
    for (int i = 1; i <= n; i++) {
      if (d[i] == Max)
        cM0++;
      else if (d[i] + 1 == Max)
        cM1++;
    }
    long long ff = 1;
    for (int i = 1; i <= n; i++) ff = ff * i % MOD;
    assert(cM0 >= 1);
    if (cM0 >= 2) {
      long long ans = 1;
      for (int i = 1; i <= n; i++) ans = ans * i % MOD;
      printf("%lld\n", ans);
    } else if (cM1 == 0) {
      printf("0\n");
    } else {
      long long ans = 1;
      for (int i = 0; i < cM1 + 1; i++) ans = ans * (n - i) % MOD;
      for (int i = 1; i <= n - cM1 - 1; i++) ans = ans * i % MOD;
      ans = ans * qpow(cM1 + 1, MOD - 2) % MOD;
      ans = (ff - ans + MOD) % MOD;
      printf("%lld\n", ans);
    }
    return;
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) solve();
    return 0;
  }
  