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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int modu = 998244353;
  long long int nofTC;
  cin >> nofTC;
  while (nofTC--) {
    long long int n;
    cin >> n;
    vector<long long int> facts(n + 1, 1);
    for (long long int i = 2; i <= n; i++) {
      facts[i] = ((i % modu) * (facts[i - 1] % modu)) % modu;
    }
    vector<long long int> tasks(n);
    for (long long int i = 0; i < n; i++) {
      cin >> tasks[i];
    }
    sort(tasks.begin(), tasks.end());
    long long int diff = tasks[n - 1] - tasks[n - 2];
    if (diff == 0) {
      cout << facts[n] << "\n";
      continue;
    } else if (diff >= 2) {
      cout << "0\n";
      continue;
    }
    long long int freqof2ndMax = 0;
    for (long long int i = 0; i < n; i++) {
      if (tasks[n - 2] == tasks[i]) {
        freqof2ndMax++;
      }
    }
    long long int res = facts[n];
    for (long long int i = freqof2ndMax; i < n; i++) {
      long long int notNice =
          ((((facts[i] % modu) * (facts[freqof2ndMax] % modu) *
             (facts[n - freqof2ndMax - 1] % modu)) %
            modu) /
           (((facts[freqof2ndMax] % modu) * (facts[i - freqof2ndMax] % modu))) %
           modu) %
          modu;
      res = ((res % modu) - (notNice % modu) + modu) % modu;
    }
    cout << res << "\n";
  }
}
