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;
long long modder = 998244353;
long long fermats_little_theorem(long long number, long long power)
{
  if(power == 0)
  {return 1;}
  long long out = fermats_little_theorem(number, power / 2);
  out = out * out % modder;
  if(power % 2 == 1)
  {out = out * number % modder;}
  return out;
}
void processor()
{
  int n; cin >> n;
  map<int, int> hmap;
  set<int, greater<int>> uniques;
  int arr[n] = {};
  for(int i = 0; i < n; i++)
  {cin >> arr[i]; hmap[arr[i]]++; uniques.insert(arr[i]);}
  int mini = *min_element(arr, arr + n);
  int maxi = *max_element(arr, arr + n);
  long long answer = 1, modder = 998244353;
  for(int val : uniques)
  {
    if(val != mini && hmap[val - 1] == 0)
    {cout << 0 << endl; return;}
  }
  if(uniques.size() == 1)
  {
    for(int i = hmap[*uniques.begin()]; i > 1; i--)
    {answer *= i; answer %= modder;}
    cout << answer << endl;
    return;
  }
  int remaining = n - hmap[maxi] - 1;
  long long intermediate = 1;
  for(int i = remaining; i > 1; i--)
  {intermediate *= i; intermediate %= modder;}
  long long last = 1, lastans = 1;
  for(int i = hmap[maxi]; i > 1; i--)
  {lastans *= i; lastans %= modder;}
  answer = lastans * intermediate % modder * hmap[maxi - 1];
  answer %= modder;
  for(int i = hmap[maxi] + 1; i < n; i++)
  {
    answer += (((lastans * intermediate) % modder) * i) % modder * fermats_little_theorem(last, modder - 2) % modder * hmap[maxi - 1] % modder;
    answer %= modder;
    lastans = lastans * i % modder * fermats_little_theorem(last++, modder - 2) % modder;
  }
  cout << (answer * (hmap[maxi - 1] > 1?fermats_little_theorem(2, modder - 2):1)) % modder << endl;
  return;
}

int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int count, counter = 0;
  cin >> count;
  while(counter++ < count)
  {processor();}
  return 0;
}
//Written by Namit Shah (Eclecticity).
//Hack/Challenge this if you wish to salvage a few points off a destined-to-sink ship.
