#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 10;
const long long mod = 998244353;
long long a[N];
long long rjc[N];
long long jc[N];
long long inverse(long long n) {
  long long ans = 1, p = 998244351;
  while (p) {
    if (p & 1) ans = (ans * n) % 998244353;
    n = (n * n) % 998244353;
    p >>= 1;
  }
  return ans;
}
void init() {
  long long res = 1;
  rjc[0] = jc[0] = 1;
  for (long long i = 1; i <= 2e5; i++) {
    jc[i] = (jc[i - 1] * i) % mod;
    rjc[i] = (rjc[i - 1] * inverse(i)) % mod;
  }
}
long long cal(long long m, long long n) {
  return ((jc[n] * rjc[n - m]) % mod) % mod;
}
signed main() {
  init();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long ans = 0, cnt1 = 0, cnt2 = 0;
    sort(a, a + n, greater<long long>());
    long long maxx = a[0];
    for (long long i = 0; i < n; i++) {
      if (a[i] == maxx) cnt1++;
      if (a[i] == maxx - 1) cnt2++;
      if (a[i] < maxx - 1) break;
    }
    if (cnt1 >= 2)
      cout << jc[n] << '\n';
    else if (cnt1 < n && cnt2 == 0)
      cout << 0 << '\n';
    else {
      long long ans = jc[n];
      ans = (ans - jc[n - 1] + mod) % mod;
      long long cnt = n - cnt1 - cnt2;
      for (long long i = 1; i <= cnt; i++) {
        long long temp = cal(i, cnt) % mod;
        temp = (temp * jc[n - i - 1]) % mod;
        ans -= temp;
        ans = (ans + mod) % mod;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}

### 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 N = 2e5 + 10;
const long long mod = 998244353;
long long a[N];
long long rjc[N];
long long jc[N];
long long inverse(long long n) {
  long long ans = 1, p = 998244351;
  while (p) {
    if (p & 1) ans = (ans * n) % 998244353;
    n = (n * n) % 998244353;
    p >>= 1;
  }
  return ans;
}
void init() {
  long long res = 1;
  rjc[0] = jc[0] = 1;
  for (long long i = 1; i <= 2e5; i++) {
    jc[i] = (jc[i - 1] * i) % mod;
    rjc[i] = (rjc[i - 1] * inverse(i)) % mod;
  }
}
long long cal(long long m, long long n) {
  return ((jc[n] * rjc[n - m]) % mod) % mod;
}
signed main() {
  init();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long ans = 0, cnt1 = 0, cnt2 = 0;
    sort(a, a + n, greater<long long>());
    long long maxx = a[0];
    for (long long i = 0; i < n; i++) {
      if (a[i] == maxx) cnt1++;
      if (a[i] == maxx - 1) cnt2++;
      if (a[i] < maxx - 1) break;
    }
    if (cnt1 >= 2)
      cout << jc[n] << '\n';
    else if (cnt1 < n && cnt2 == 0)
      cout << 0 << '\n';
    else {
      long long ans = jc[n];
      ans = (ans - jc[n - 1] + mod) % mod;
      long long cnt = n - cnt1 - cnt2;
      for (long long i = 1; i <= cnt; i++) {
        long long temp = cal(i, cnt) % mod;
        temp = (temp * jc[n - i - 1]) % mod;
        ans -= temp;
        ans = (ans + mod) % mod;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}

```