#include <bits/stdc++.h>
using namespace std;
long long fact[300005];
long long a[300005];
long long bigmod(long long x, long long p) {
  if (p == 0) return 1LL;
  long long y = bigmod(x, p / 2);
  if (p % 2 == 1) {
    return ((((y % 998244353) * (y % 998244353)) % 998244353) *
            (x % 998244353)) %
           998244353;
  }
  return (y % 998244353) * (y % 998244353) % 998244353;
}
long long invmod(long long x) { return bigmod(x, 998244353 - 2); }
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long large = a[n - 1];
  long long cntL = 0, cntl = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == large)
      cntL++;
    else if (a[i] == large - 1)
      cntl++;
  }
  if (cntL > 1) {
    cout << fact[n] << endl;
    return;
  }
  if (cntl == 0) {
    cout << 0 << endl;
    return;
  }
  cout << ((fact[n] -
            (((fact[n] * fact[cntl]) % 998244353) * invmod(fact[cntl + 1])) %
                998244353) %
               998244353 +
           998244353) %
              998244353
       << endl;
}
int main() {
  int t;
  cin >> t;
  fact[0] = 1;
  for (long long i = 1; i <= 200000; i++)
    fact[i] = (fact[i - 1] * i) % 998244353;
  while (t--) {
    solve();
  }
}

### 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;
long long fact[300005];
long long a[300005];
long long bigmod(long long x, long long p) {
  if (p == 0) return 1LL;
  long long y = bigmod(x, p / 2);
  if (p % 2 == 1) {
    return ((((y % 998244353) * (y % 998244353)) % 998244353) *
            (x % 998244353)) %
           998244353;
  }
  return (y % 998244353) * (y % 998244353) % 998244353;
}
long long invmod(long long x) { return bigmod(x, 998244353 - 2); }
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long large = a[n - 1];
  long long cntL = 0, cntl = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == large)
      cntL++;
    else if (a[i] == large - 1)
      cntl++;
  }
  if (cntL > 1) {
    cout << fact[n] << endl;
    return;
  }
  if (cntl == 0) {
    cout << 0 << endl;
    return;
  }
  cout << ((fact[n] -
            (((fact[n] * fact[cntl]) % 998244353) * invmod(fact[cntl + 1])) %
                998244353) %
               998244353 +
           998244353) %
              998244353
       << endl;
}
int main() {
  int t;
  cin >> t;
  fact[0] = 1;
  for (long long i = 1; i <= 200000; i++)
    fact[i] = (fact[i - 1] * i) % 998244353;
  while (t--) {
    solve();
  }
}

```