#include <bits/stdc++.h>
using namespace std;
int dx[4] = {0, 0, -1, 1}, dy[4] = {1, -1, 0, 0};
long long fact[200005];
void comp() {
  fact[0] = 1;
  for (int i = 0; i < 200005 - 1; i++) {
    fact[i + 1] = ((i + 1) * fact[i]) % 998244353;
  }
}
long long exp(long long n, long long x) {
  if (x == 0) return 1;
  if (x == 1) return n;
  long long tmp = exp(n, x / 2);
  tmp *= tmp;
  tmp %= 998244353;
  if (x % 2 == 1) tmp *= n;
  return tmp % 998244353;
}
int main() {
  ios::sync_with_stdio(0);
  int t;
  cin >> t;
  comp();
  for (int q = 0; q < t; q++) {
    int n;
    cin >> n;
    vector<int> arr;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      arr.push_back(x);
    }
    sort(arr.begin(), arr.end(), greater<int>());
    if (arr[0] == arr[1] + 1) {
      int cnt = 1;
      while (cnt < n && arr[cnt] == arr[1]) cnt++;
      cnt--;
      cout << ((fact[n] * cnt) % 998244353 * exp(cnt + 1, 998244353 - 2)) %
                  998244353
           << '\n';
    } else if (arr[0] == arr[1])
      cout << fact[n] << '\n';
    else
      cout << "0\n";
  }
}

### Prompt

Generate 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;
int dx[4] = {0, 0, -1, 1}, dy[4] = {1, -1, 0, 0};
long long fact[200005];
void comp() {
  fact[0] = 1;
  for (int i = 0; i < 200005 - 1; i++) {
    fact[i + 1] = ((i + 1) * fact[i]) % 998244353;
  }
}
long long exp(long long n, long long x) {
  if (x == 0) return 1;
  if (x == 1) return n;
  long long tmp = exp(n, x / 2);
  tmp *= tmp;
  tmp %= 998244353;
  if (x % 2 == 1) tmp *= n;
  return tmp % 998244353;
}
int main() {
  ios::sync_with_stdio(0);
  int t;
  cin >> t;
  comp();
  for (int q = 0; q < t; q++) {
    int n;
    cin >> n;
    vector<int> arr;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      arr.push_back(x);
    }
    sort(arr.begin(), arr.end(), greater<int>());
    if (arr[0] == arr[1] + 1) {
      int cnt = 1;
      while (cnt < n && arr[cnt] == arr[1]) cnt++;
      cnt--;
      cout << ((fact[n] * cnt) % 998244353 * exp(cnt + 1, 998244353 - 2)) %
                  998244353
           << '\n';
    } else if (arr[0] == arr[1])
      cout << fact[n] << '\n';
    else
      cout << "0\n";
  }
}

```