#include <bits/stdc++.h>
using namespace std;
inline void file_handling() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
void solve() {
  long long N;
  cin >> N;
  vector<long long> v(N);
  for (long long &i : v) cin >> i;
  long long mx = *max_element(v.begin(), v.end());
  long long count_mx = count(v.begin(), v.end(), mx);
  long long k = count(v.begin(), v.end(), mx - 1);
  long long fac1 = 1LL;
  long long fac2 = 1LL;
  for (long long i = 1; i <= N; ++i) {
    fac1 = ((fac1 % 998244353) * i) % 998244353;
    if (i != k + 1) fac2 = ((fac2 % 998244353) * i) % 998244353;
  }
  if (count_mx > 1) {
    cout << fac1 << '\n';
  } else if (k == 0) {
    cout << 0 << '\n';
  } else {
    cout << (fac1 - fac2 + 998244353) % 998244353 << '\n';
  }
}
signed main() {
  file_handling();
  long long testCases = 1;
  cin >> testCases;
  while (testCases--) solve();
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline void file_handling() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
void solve() {
  long long N;
  cin >> N;
  vector<long long> v(N);
  for (long long &i : v) cin >> i;
  long long mx = *max_element(v.begin(), v.end());
  long long count_mx = count(v.begin(), v.end(), mx);
  long long k = count(v.begin(), v.end(), mx - 1);
  long long fac1 = 1LL;
  long long fac2 = 1LL;
  for (long long i = 1; i <= N; ++i) {
    fac1 = ((fac1 % 998244353) * i) % 998244353;
    if (i != k + 1) fac2 = ((fac2 % 998244353) * i) % 998244353;
  }
  if (count_mx > 1) {
    cout << fac1 << '\n';
  } else if (k == 0) {
    cout << 0 << '\n';
  } else {
    cout << (fac1 - fac2 + 998244353) % 998244353 << '\n';
  }
}
signed main() {
  file_handling();
  long long testCases = 1;
  cin >> testCases;
  while (testCases--) solve();
  return 0;
}

```