#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
int T, N;
int A[200001];
long long Fact[200001], Inv_Fact[200001];
long long my_pow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return ret;
}
void solve() {
  int Max = 0, max_cnt = 0, almost_max_cnt = 0;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    Max = max(Max, A[i]);
  }
  for (int i = 1; i <= N; i++) {
    if (A[i] == Max) max_cnt++;
  }
  if (max_cnt > 1) {
    cout << Fact[N] << "\n";
    return;
  }
  for (int i = 1; i <= N; i++) {
    if (A[i] == Max - 1) almost_max_cnt++;
  }
  if (!almost_max_cnt) {
    cout << "0\n";
    return;
  }
  long long ans = (Fact[N] - Fact[almost_max_cnt] * Fact[N] % MOD *
                                 Inv_Fact[almost_max_cnt + 1] % MOD) %
                  MOD;
  if (ans < 0) ans += MOD;
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  Fact[0] = 1;
  for (int i = 1; i <= 200000; i++) Fact[i] = (Fact[i - 1] * i) % MOD;
  Inv_Fact[200000] = my_pow(Fact[200000], MOD - 2);
  for (int i = 199999; i >= 0; i--)
    Inv_Fact[i] = Inv_Fact[i + 1] * (i + 1) % MOD;
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MOD = 998244353;
int T, N;
int A[200001];
long long Fact[200001], Inv_Fact[200001];
long long my_pow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return ret;
}
void solve() {
  int Max = 0, max_cnt = 0, almost_max_cnt = 0;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    Max = max(Max, A[i]);
  }
  for (int i = 1; i <= N; i++) {
    if (A[i] == Max) max_cnt++;
  }
  if (max_cnt > 1) {
    cout << Fact[N] << "\n";
    return;
  }
  for (int i = 1; i <= N; i++) {
    if (A[i] == Max - 1) almost_max_cnt++;
  }
  if (!almost_max_cnt) {
    cout << "0\n";
    return;
  }
  long long ans = (Fact[N] - Fact[almost_max_cnt] * Fact[N] % MOD *
                                 Inv_Fact[almost_max_cnt + 1] % MOD) %
                  MOD;
  if (ans < 0) ans += MOD;
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  Fact[0] = 1;
  for (int i = 1; i <= 200000; i++) Fact[i] = (Fact[i - 1] * i) % MOD;
  Inv_Fact[200000] = my_pow(Fact[200000], MOD - 2);
  for (int i = 199999; i >= 0; i--)
    Inv_Fact[i] = Inv_Fact[i + 1] * (i + 1) % MOD;
  cin >> T;
  while (T--) solve();
  return 0;
}

```