#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 20;
const long long mod = 998244353;
long long fact[N];
map<long long, long long> vis;
long long ksm(long long a, long long b) {
  if (b == 0) return 1;
  if (b & 1)
    return a * ksm(a, b - 1) % mod;
  else {
    long long k = ksm(a, b / 2);
    return k * k % mod;
  }
}
long long C(long long m, long long n) {
  if (m == 0) return 1;
  return fact[n] * ksm(fact[m] * fact[n - m] % mod, mod - 2) % mod;
}
signed main() {
  fact[0] = 1;
  for (long long i = 1; i < N; i++) {
    fact[i] = fact[i - 1] * i % mod;
  }
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    vis.clear();
    long long mx = 0;
    for (long long i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      vis[x]++;
      mx = max(x, mx);
    }
    if (vis[mx] > 1)
      cout << fact[n] << endl;
    else {
      if (vis[mx - 1] == 0) {
        cout << 0 << endl;
      } else {
        long long all = n - vis[mx] - vis[mx - 1];
        long long res = fact[n];
        for (long long i = vis[mx - 1] + 1; i <= n; i++) {
          res = ((res - C(n - i, all) * fact[i - 1] % mod * fact[n - i] % mod) +
                 mod) %
                mod;
        }
        cout << res << endl;
      }
    }
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 N = 1e6 + 20;
const long long mod = 998244353;
long long fact[N];
map<long long, long long> vis;
long long ksm(long long a, long long b) {
  if (b == 0) return 1;
  if (b & 1)
    return a * ksm(a, b - 1) % mod;
  else {
    long long k = ksm(a, b / 2);
    return k * k % mod;
  }
}
long long C(long long m, long long n) {
  if (m == 0) return 1;
  return fact[n] * ksm(fact[m] * fact[n - m] % mod, mod - 2) % mod;
}
signed main() {
  fact[0] = 1;
  for (long long i = 1; i < N; i++) {
    fact[i] = fact[i - 1] * i % mod;
  }
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    vis.clear();
    long long mx = 0;
    for (long long i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      vis[x]++;
      mx = max(x, mx);
    }
    if (vis[mx] > 1)
      cout << fact[n] << endl;
    else {
      if (vis[mx - 1] == 0) {
        cout << 0 << endl;
      } else {
        long long all = n - vis[mx] - vis[mx - 1];
        long long res = fact[n];
        for (long long i = vis[mx - 1] + 1; i <= n; i++) {
          res = ((res - C(n - i, all) * fact[i - 1] % mod * fact[n - i] % mod) +
                 mod) %
                mod;
        }
        cout << res << endl;
      }
    }
  }
}

```