#include <bits/stdc++.h>
using namespace std;
long long f[200005];
int a[200005], n;
long long getmod(long long x) {
  long long res = 1;
  int t = 998244353 - 2;
  for (; t; t /= 2, x = (x * x) % 998244353)
    if (t & 1) res = (res * x) % 998244353;
  return res;
}
long long C(int k, int n) {
  long long res = f[n];
  res = (res * getmod(f[k])) % 998244353;
  res = (res * getmod(f[n - k])) % 998244353;
  return res;
}
void Solve() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  sort(a + 1, a + n + 1, greater<int>());
  if (a[1] == a[2]) {
    cout << f[n] << '\n';
    return;
  }
  if (a[1] - a[2] > 1) {
    cout << 0 << '\n';
    return;
  }
  int cnt = 0;
  for (int i = 2; i <= n; ++i)
    if (a[i] == a[2]) ++cnt;
  long long ans = 0;
  for (int i = 1; i < n; ++i) {
    if (cnt >= i)
      ans = (ans + f[n - 1]) % 998244353;
    else {
      ans = (ans + (f[n - 1] -
                    ((f[cnt] * C(cnt, i - 1) % 998244353) * f[n - cnt - 1] %
                     998244353) +
                    998244353)) %
            998244353;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  if (fopen("D:\\C++\\clion\\"
            ".inp",
            "r")) {
    freopen(
        "D:\\C++\\clion\\"
        ".inp",
        "r", stdin);
    freopen(
        "D:\\C++\\clion\\"
        ".out",
        "w", stdout);
  }
  f[0] = 1;
  for (int i = 1; i < 200005; ++i) f[i] = f[i - 1] * i % 998244353;
  int t;
  cin >> t;
  while (t--) {
    Solve();
  }
  cerr << "Time collapse : " << fixed << setprecision(3)
       << 1.000 * clock() / CLOCKS_PER_SEC;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 f[200005];
int a[200005], n;
long long getmod(long long x) {
  long long res = 1;
  int t = 998244353 - 2;
  for (; t; t /= 2, x = (x * x) % 998244353)
    if (t & 1) res = (res * x) % 998244353;
  return res;
}
long long C(int k, int n) {
  long long res = f[n];
  res = (res * getmod(f[k])) % 998244353;
  res = (res * getmod(f[n - k])) % 998244353;
  return res;
}
void Solve() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  sort(a + 1, a + n + 1, greater<int>());
  if (a[1] == a[2]) {
    cout << f[n] << '\n';
    return;
  }
  if (a[1] - a[2] > 1) {
    cout << 0 << '\n';
    return;
  }
  int cnt = 0;
  for (int i = 2; i <= n; ++i)
    if (a[i] == a[2]) ++cnt;
  long long ans = 0;
  for (int i = 1; i < n; ++i) {
    if (cnt >= i)
      ans = (ans + f[n - 1]) % 998244353;
    else {
      ans = (ans + (f[n - 1] -
                    ((f[cnt] * C(cnt, i - 1) % 998244353) * f[n - cnt - 1] %
                     998244353) +
                    998244353)) %
            998244353;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  if (fopen("D:\\C++\\clion\\"
            ".inp",
            "r")) {
    freopen(
        "D:\\C++\\clion\\"
        ".inp",
        "r", stdin);
    freopen(
        "D:\\C++\\clion\\"
        ".out",
        "w", stdout);
  }
  f[0] = 1;
  for (int i = 1; i < 200005; ++i) f[i] = f[i - 1] * i % 998244353;
  int t;
  cin >> t;
  while (t--) {
    Solve();
  }
  cerr << "Time collapse : " << fixed << setprecision(3)
       << 1.000 * clock() / CLOCKS_PER_SEC;
  return 0;
}

```