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 mxN = 2e5 + 7;
const long long MOD = 998244353;
int a[mxN];
long long silnia[mxN];
int pref[mxN];
void solve() {
  int n;
  cin >> n;
  map<int, int> cnt;
  int mx = -1;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; i++) mx = max(mx, a[i]);
  for (long long i = 1; i <= n; i++) cnt[a[i]]++;
  if (cnt[mx] > 1)
    cout << silnia[n] << "\n";
  else if (a[n] - a[n - 1] > 1)
    cout << "0\n";
  else {
    int ile = cnt[a[n - 1]];
    long long res = pref[n - 1];
    res *= ile;
    res %= MOD;
    res *= (silnia[n - 2] / (silnia[ile - 1] * silnia[n - 2 - ile + 1]));
    res %= MOD;
    res *= silnia[n - ile - 1];
    res %= MOD;
    cout << res << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  silnia[0] = 1;
  for (long long i = 1; i <= 2e5; i++) silnia[i] = (silnia[i - 1] * i) % MOD;
  pref[0] = 0;
  for (long long i = 1; i <= 2e5; i++) pref[i] = pref[i - 1] + i;
  int t;
  cin >> t;
  while (t--) solve();
}
