#include <bits/stdc++.h>
using namespace std;
const long long C1 = 1e9 + 7;
const long long C2 = 998244353;
void yes() { cout << "YES\n"; }
void no() { cout << "NO\n"; }
long long binpow(long long a, long long b, long long m = C1) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
vector<long long> fac(200001);
vector<long long> invfac(200001);
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  long long cnt{};
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long m = *max_element(v.begin(), v.end());
  map<long long, long long> freq;
  for (long long i = 0; i < n; i++) {
    if (m == v[i]) cnt++;
    freq[v[i]]++;
  }
  if (cnt == 1) {
    if (freq[m - 1] != 0) {
      long long x = freq[m - 1];
      long long ans = fac[n];
      ans -= ((fac[n] * binpow(x + 1, C2 - 2, C2)) % C2);
      ans += C2;
      ans %= C2;
      cout << ans << '\n';
    } else {
      cout << 0 << '\n';
    }
  } else {
    long long ans = fac[n];
    cout << ans << '\n';
  }
}
signed main() {
  long long t = 1;
  cin >> t;
  fac[0] = 1;
  fac[1] = 1;
  for (long long i = 2; i <= 2e5; i++) {
    fac[i] = (fac[i - 1] * i) % C2;
  }
  invfac[200000] = binpow(fac[200000], C2 - 2, C2);
  for (long long i = 200000 - 1; i >= 1; i--) {
    invfac[i] = (invfac[i + 1] * (i + 1)) % C2;
  }
  while (t--) {
    solve();
  }
}

### Prompt

Generate a CPP solution to 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 C1 = 1e9 + 7;
const long long C2 = 998244353;
void yes() { cout << "YES\n"; }
void no() { cout << "NO\n"; }
long long binpow(long long a, long long b, long long m = C1) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
vector<long long> fac(200001);
vector<long long> invfac(200001);
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  long long cnt{};
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long m = *max_element(v.begin(), v.end());
  map<long long, long long> freq;
  for (long long i = 0; i < n; i++) {
    if (m == v[i]) cnt++;
    freq[v[i]]++;
  }
  if (cnt == 1) {
    if (freq[m - 1] != 0) {
      long long x = freq[m - 1];
      long long ans = fac[n];
      ans -= ((fac[n] * binpow(x + 1, C2 - 2, C2)) % C2);
      ans += C2;
      ans %= C2;
      cout << ans << '\n';
    } else {
      cout << 0 << '\n';
    }
  } else {
    long long ans = fac[n];
    cout << ans << '\n';
  }
}
signed main() {
  long long t = 1;
  cin >> t;
  fac[0] = 1;
  fac[1] = 1;
  for (long long i = 2; i <= 2e5; i++) {
    fac[i] = (fac[i - 1] * i) % C2;
  }
  invfac[200000] = binpow(fac[200000], C2 - 2, C2);
  for (long long i = 200000 - 1; i >= 1; i--) {
    invfac[i] = (invfac[i + 1] * (i + 1)) % C2;
  }
  while (t--) {
    solve();
  }
}

```