#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 100;
const long long mod = 998244353;
long long n, m, a[maxn], _;
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  long long ma = *max_element(a + 1, a + 1 + n);
  long long num = count(a + 1, a + 1 + n, ma),
            k = count(a + 1, a + 1 + n, ma - 1);
  long long res = 1, tmp = 1;
  for (long long i = 1; i <= n; i++) {
    if (i != k + 1) tmp = tmp * i % mod;
    res = res * i % mod;
  }
  if (num == 1) res = (res - tmp + mod) % mod;
  cout << res << endl;
}
signed main() {
  long long t;
  cin >> t;
  while (t--) solve();
}

### 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;
const long long maxn = 2e5 + 100;
const long long mod = 998244353;
long long n, m, a[maxn], _;
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  long long ma = *max_element(a + 1, a + 1 + n);
  long long num = count(a + 1, a + 1 + n, ma),
            k = count(a + 1, a + 1 + n, ma - 1);
  long long res = 1, tmp = 1;
  for (long long i = 1; i <= n; i++) {
    if (i != k + 1) tmp = tmp * i % mod;
    res = res * i % mod;
  }
  if (num == 1) res = (res - tmp + mod) % mod;
  cout << res << endl;
}
signed main() {
  long long t;
  cin >> t;
  while (t--) solve();
}

```