#include <bits/stdc++.h>
using namespace std;
const long long p = 998244353;
long long shu[200010];
long long fact[200010];
long long infact[200010];
long long qmi(long long a, long long b) {
  long long ans = 1;
  long long base = a;
  while (b) {
    if (b & 1) ans = base * ans % p;
    base = base * base % p;
    b >>= 1;
  }
  return ans;
}
long long ni(long long a) { return qmi(a, p - 2); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int T;
  cin >> T;
  fact[0] = 1;
  infact[0] = 1;
  for (int i = 1; i <= 200005; i++) {
    fact[i] = fact[i - 1] * i % p;
  }
  while (T--) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> shu[i];
    sort(shu, shu + n);
    int t = shu[n - 1] - shu[n - 2];
    if (t == 1) {
      long long s = count(shu, shu + n, shu[n - 2]);
      s++;
      infact[s] = ni(fact[s]);
      long long tt = ((fact[s] - fact[s - 1]) % p + p) % p;
      printf("%lld\n", fact[n] * tt % p * infact[s] % p);
    }
    if (t >= 2) printf("0\n");
    if (t == 0) printf("%lld\n", fact[n]);
  }
}

### 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;
const long long p = 998244353;
long long shu[200010];
long long fact[200010];
long long infact[200010];
long long qmi(long long a, long long b) {
  long long ans = 1;
  long long base = a;
  while (b) {
    if (b & 1) ans = base * ans % p;
    base = base * base % p;
    b >>= 1;
  }
  return ans;
}
long long ni(long long a) { return qmi(a, p - 2); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int T;
  cin >> T;
  fact[0] = 1;
  infact[0] = 1;
  for (int i = 1; i <= 200005; i++) {
    fact[i] = fact[i - 1] * i % p;
  }
  while (T--) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> shu[i];
    sort(shu, shu + n);
    int t = shu[n - 1] - shu[n - 2];
    if (t == 1) {
      long long s = count(shu, shu + n, shu[n - 2]);
      s++;
      infact[s] = ni(fact[s]);
      long long tt = ((fact[s] - fact[s - 1]) % p + p) % p;
      printf("%lld\n", fact[n] * tt % p * infact[s] % p);
    }
    if (t >= 2) printf("0\n");
    if (t == 0) printf("%lld\n", fact[n]);
  }
}

```