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.
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const int N = 2e5 + 100;
int n;
int arr[N];
long long fac[N], inv[N];
long long quickmod(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
void init() {
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= N; i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = quickmod(fac[i], mod - 2);
  }
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  sort(arr + 1, arr + 1 + n, greater<int>());
  int a = arr[1], b = arr[2];
  long long ans = fac[n];
  if (a - b > 1)
    ans = 0;
  else if (a == b + 1) {
    int cnt = 1;
    for (int i = 3; i <= n; i++)
      if (arr[i] == b)
        cnt++;
      else if (arr[i] < b)
        break;
    long long res = fac[cnt] * ans % mod;
    res = res * inv[n - cnt + 1] % mod;
    ans = (ans - res + mod) % mod;
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();
  int Case;
  cin >> Case;
  while (Case--) solve();
  return 0;
}
