#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
long long fac[200010], infact[200010];
long long qmi(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = res * a % mod;
    }
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
void init() {
  fac[0] = 1;
  infact[0] = 1;
  for (int i = 1; i <= 200000; i++) {
    fac[i] = i * fac[i - 1] % mod;
    infact[i] = qmi(fac[i], mod - 2) % mod;
  }
}
int main() {
  int t;
  cin >> t;
  init();
  while (t--) {
    int fir[2] = {0}, sec[2] = {0};
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      int num;
      cin >> num;
      if (sec[0] == num) {
        sec[1]++;
      } else if (fir[0] == num) {
        fir[1]++;
      } else if (fir[0] < num) {
        sec[0] = fir[0];
        sec[1] = fir[1];
        fir[0] = num;
        fir[1] = 1;
      } else if (num > sec[0]) {
        sec[0] = num;
        sec[1] = 1;
      }
    }
    if (fir[1] > 1) {
      cout << fac[n] << endl;
      continue;
    } else if (fir[0] - sec[0] >= 2) {
      cout << 0 << endl;
      continue;
    } else {
      long long ans;
      ans = (fac[n] * infact[n - sec[1] - 1] % mod * infact[sec[1] + 1] % mod +
             mod) %
            mod;
      int k =
          fac[sec[1]] * (sec[1]) % mod * fac[n - sec[1] - 1] % mod * ans % mod;
      cout << k % mod << endl;
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 int mod = 998244353;
long long fac[200010], infact[200010];
long long qmi(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = res * a % mod;
    }
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
void init() {
  fac[0] = 1;
  infact[0] = 1;
  for (int i = 1; i <= 200000; i++) {
    fac[i] = i * fac[i - 1] % mod;
    infact[i] = qmi(fac[i], mod - 2) % mod;
  }
}
int main() {
  int t;
  cin >> t;
  init();
  while (t--) {
    int fir[2] = {0}, sec[2] = {0};
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      int num;
      cin >> num;
      if (sec[0] == num) {
        sec[1]++;
      } else if (fir[0] == num) {
        fir[1]++;
      } else if (fir[0] < num) {
        sec[0] = fir[0];
        sec[1] = fir[1];
        fir[0] = num;
        fir[1] = 1;
      } else if (num > sec[0]) {
        sec[0] = num;
        sec[1] = 1;
      }
    }
    if (fir[1] > 1) {
      cout << fac[n] << endl;
      continue;
    } else if (fir[0] - sec[0] >= 2) {
      cout << 0 << endl;
      continue;
    } else {
      long long ans;
      ans = (fac[n] * infact[n - sec[1] - 1] % mod * infact[sec[1] + 1] % mod +
             mod) %
            mod;
      int k =
          fac[sec[1]] * (sec[1]) % mod * fac[n - sec[1] - 1] % mod * ans % mod;
      cout << k % mod << endl;
    }
  }
  return 0;
}

```