#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
const long long mod = 998244353;
long long f[N], num[N];
long long ppow(long long x, int i) {
  long long ans = 1, res = x % mod;
  while (i) {
    if (i & 1) ans = (res * ans) % mod;
    res = (res * res) % mod;
    i >>= 1;
  }
  return ans;
}
void init() {
  f[0] = 1;
  for (int i = 1; i < N; i++) {
    f[i] = (f[i - 1] * i) % mod;
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> num[i];
  sort(num + 1, num + 1 + n);
  if (num[n] == num[n - 1]) {
    cout << f[n] << endl;
  } else {
    if (num[n] != (num[n - 1] + 1)) {
      cout << 0 << endl;
      return;
    }
    int cnt = 1;
    for (int i = n - 1; i >= 2; i--) {
      if (num[i - 1] == num[i])
        cnt++;
      else
        break;
    }
    cnt++;
    long long ans = f[n] * ppow(cnt, mod - 2) % mod;
    ans = (f[n] - ans + mod) % mod;
    cout << ans << endl;
  }
}
int main() {
  init();
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}

### Prompt

Create a solution in Cpp for 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 int N = 1e6;
const long long mod = 998244353;
long long f[N], num[N];
long long ppow(long long x, int i) {
  long long ans = 1, res = x % mod;
  while (i) {
    if (i & 1) ans = (res * ans) % mod;
    res = (res * res) % mod;
    i >>= 1;
  }
  return ans;
}
void init() {
  f[0] = 1;
  for (int i = 1; i < N; i++) {
    f[i] = (f[i - 1] * i) % mod;
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> num[i];
  sort(num + 1, num + 1 + n);
  if (num[n] == num[n - 1]) {
    cout << f[n] << endl;
  } else {
    if (num[n] != (num[n - 1] + 1)) {
      cout << 0 << endl;
      return;
    }
    int cnt = 1;
    for (int i = n - 1; i >= 2; i--) {
      if (num[i - 1] == num[i])
        cnt++;
      else
        break;
    }
    cnt++;
    long long ans = f[n] * ppow(cnt, mod - 2) % mod;
    ans = (f[n] - ans + mod) % mod;
    cout << ans << endl;
  }
}
int main() {
  init();
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}

```