#include <bits/stdc++.h>
using namespace std;
const int si = 200001;
long long dp[si];
long long power(long long x, int y, int p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, int p) { return power(n, p - 2, p); }
long long nCrModPFermat(long long n, int r, int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (dp[n] * modInverse(dp[r], p) % p * modInverse(dp[n - r], p) % p) % p;
}
void solve() {
  long long n;
  cin >> n;
  int arr[n];
  int flag = true;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  if ((arr[n - 1] == arr[n - 2]) > 1) {
    cout << 0;
    return;
  }
  if (arr[n - 1] == arr[n - 2]) {
    long long ans = dp[n];
    ans = ans % 998244353;
    cout << ans;
    return;
  }
  int ct = 0;
  for (int i = 0; i < n; i++) {
    if ((arr[n - 1] - arr[i]) > 1) {
      ct++;
    }
  }
  if (ct == 0) {
    long long ans = dp[n - 1] * (n - 1);
    ans = ans % 998244353;
    cout << ans;
    return;
  }
  long long ans = dp[n - 1] * (n - 1);
  ans = ans % 998244353;
  for (int r = 1; r <= ct; r++) {
    long long inv = nCrModPFermat(ct, r, 998244353);
    long long ann = (((inv * dp[r]) % 998244353) * dp[n - r - 1]) % 998244353;
    ans = ans - ann + 998244353;
    ans = ans % 998244353;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  dp[0] = 1;
  for (int i = 1; i < si; i++) {
    dp[i] = dp[i - 1] * i;
    dp[i] = dp[i] % 998244353;
  }
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
    cout << endl;
  }
  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 int si = 200001;
long long dp[si];
long long power(long long x, int y, int p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, int p) { return power(n, p - 2, p); }
long long nCrModPFermat(long long n, int r, int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (dp[n] * modInverse(dp[r], p) % p * modInverse(dp[n - r], p) % p) % p;
}
void solve() {
  long long n;
  cin >> n;
  int arr[n];
  int flag = true;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  if ((arr[n - 1] == arr[n - 2]) > 1) {
    cout << 0;
    return;
  }
  if (arr[n - 1] == arr[n - 2]) {
    long long ans = dp[n];
    ans = ans % 998244353;
    cout << ans;
    return;
  }
  int ct = 0;
  for (int i = 0; i < n; i++) {
    if ((arr[n - 1] - arr[i]) > 1) {
      ct++;
    }
  }
  if (ct == 0) {
    long long ans = dp[n - 1] * (n - 1);
    ans = ans % 998244353;
    cout << ans;
    return;
  }
  long long ans = dp[n - 1] * (n - 1);
  ans = ans % 998244353;
  for (int r = 1; r <= ct; r++) {
    long long inv = nCrModPFermat(ct, r, 998244353);
    long long ann = (((inv * dp[r]) % 998244353) * dp[n - r - 1]) % 998244353;
    ans = ans - ann + 998244353;
    ans = ans % 998244353;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  dp[0] = 1;
  for (int i = 1; i < si; i++) {
    dp[i] = dp[i - 1] * i;
    dp[i] = dp[i] % 998244353;
  }
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
    cout << endl;
  }
  return 0;
}

```