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;
  int yh() {
    int ret = 0;
    bool f = 0;
    char c = getchar();
    while (!isdigit(c)) {
      if (c == EOF) return -1;
      if (c == '-') f = 1;
      c = getchar();
    }
    while (isdigit(c)) ret = (ret << 3) + (ret << 1) + (c ^ 48), c = getchar();
    return f ? -ret : ret;
  }
  const int maxn = 3e5 + 5, mod = 998244353;
  int n, m;
  int a[maxn];
  int fac[maxn], ifac[maxn];
  int mul(int x, int y) { return (long long)x * y % mod; }
  int ksm(int x, int k) {
    int ans = 1;
    for (; k; k >>= 1, x = mul(x, x))
      if (k & 1) ans = mul(ans, x);
    return ans;
  }
  int C(int x, int y) {
    if (x < y) return 0;
    return mul(fac[x], mul(ifac[y], ifac[x - y]));
  }
  int main() {
    fac[0] = 1;
    for (int i = (1), iss = (2e5); i <= iss; i++) fac[i] = mul(fac[i - 1], i);
    ifac[200000] = ksm(fac[200000], mod - 2);
    for (int i = (199999), iss = (0); i >= iss; i--)
      ifac[i] = mul(ifac[i + 1], i + 1);
    for (int _ = (yh()), _ss = (1); _ >= _ss; _--) {
      n = yh();
      map<int, int> cnt;
      for (int i = (1), iss = (n); i <= iss; i++) a[i] = yh(), cnt[a[i]]++;
      sort(a + 1, a + 1 + n);
      int ans = fac[n];
      if (cnt[a[n]] > 1) {
        cout << ans << '\n';
        continue;
      }
      if (a[n] - a[n - 1] > 1) {
        cout << 0 << '\n';
        continue;
      }
      int oc = cnt[a[n - 1]];
      cout << (ans - mul(C(n, n - oc - 1), mul(fac[oc], fac[n - oc - 1])) + mod) %
                  mod
           << '\n';
    }
    return 0;
  }
  