#include <bits/stdc++.h>
using namespace std;
int w[200005];
const int mod = 998244353;
signed main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &w[i]);
    }
    sort(w + 1, w + n + 1);
    if (w[n] == w[n - 1]) {
      long long ans = 1;
      for (long long i = 2; i <= n; i++) {
        ans *= i;
        ans %= mod;
      }
      printf("%lld\n", ans);
      continue;
    }
    if (w[n] - w[n - 1] != 1) {
      puts("0");
      continue;
    }
    int pp = n - 1;
    while (w[pp] == w[n]) pp--;
    long long op1 = 0;
    for (int i = 1; i <= n; i++) {
      if (w[i] == w[pp]) {
        op1++;
      }
    }
    long long ans = op1;
    for (long long i = 2; i <= op1; i++) {
      ans *= i;
      ans %= mod;
    }
    for (long long i = 1; i <= (n - op1 - 1); i++) {
      ans *= (op1 + 2 + i - 1);
      ans %= mod;
    }
    printf("%lld\n", ans);
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
int w[200005];
const int mod = 998244353;
signed main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &w[i]);
    }
    sort(w + 1, w + n + 1);
    if (w[n] == w[n - 1]) {
      long long ans = 1;
      for (long long i = 2; i <= n; i++) {
        ans *= i;
        ans %= mod;
      }
      printf("%lld\n", ans);
      continue;
    }
    if (w[n] - w[n - 1] != 1) {
      puts("0");
      continue;
    }
    int pp = n - 1;
    while (w[pp] == w[n]) pp--;
    long long op1 = 0;
    for (int i = 1; i <= n; i++) {
      if (w[i] == w[pp]) {
        op1++;
      }
    }
    long long ans = op1;
    for (long long i = 2; i <= op1; i++) {
      ans *= i;
      ans %= mod;
    }
    for (long long i = 1; i <= (n - op1 - 1); i++) {
      ans *= (op1 + 2 + i - 1);
      ans %= mod;
    }
    printf("%lld\n", ans);
  }
}

```