#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
using namespace std;
template <typename val1>
void __f(const char* name, val1&& value) {
  cout << name << " = " << value << endl;
}
template <typename val1, typename... values>
void __f(const char* names, val1&& value, values&&... multiplevalue) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " = " << value << " ";
  __f(comma + 1, multiplevalue...);
}
const long long N = 2e5 + 5;
vector<long long> fac(N, 1);
long long power(long long x, int y, int p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, int p) { return power(n, p - 2, p); }
long long nCr(long long n, long long r, long long p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  const long long mod = 998244353;
  for (long long i = 2; i <= 2e5; i++) fac[i] = (fac[i - 1] * i) % mod;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int arr[n];
    for (int i = 0; i < n; i++) cin >> arr[i];
    sort(arr, arr + n);
    map<int, int> ok;
    for (int i = 0; i < n; i++) ok[arr[i]]++;
    if (ok[arr[n - 1]] > 1) {
      cout << fac[n] << endl;
    } else if (ok[arr[n - 1] - 1] == 0) {
      cout << 0 << endl;
    } else {
      long long minus = 0;
      for (int i = 0; i < n; i++) {
        int left = n - ok[arr[n - 1] - 1] - 1;
        int have = ok[arr[n - 1] - 1];
        if (i < have) continue;
        minus =
            minus + (((nCr(i, have, mod) * fac[have]) % mod) * fac[left]) % mod;
        minus %= mod;
      }
      cout << (fac[n] - minus + mod) % mod << endl;
    }
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
using namespace std;
template <typename val1>
void __f(const char* name, val1&& value) {
  cout << name << " = " << value << endl;
}
template <typename val1, typename... values>
void __f(const char* names, val1&& value, values&&... multiplevalue) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " = " << value << " ";
  __f(comma + 1, multiplevalue...);
}
const long long N = 2e5 + 5;
vector<long long> fac(N, 1);
long long power(long long x, int y, int p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, int p) { return power(n, p - 2, p); }
long long nCr(long long n, long long r, long long p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  const long long mod = 998244353;
  for (long long i = 2; i <= 2e5; i++) fac[i] = (fac[i - 1] * i) % mod;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int arr[n];
    for (int i = 0; i < n; i++) cin >> arr[i];
    sort(arr, arr + n);
    map<int, int> ok;
    for (int i = 0; i < n; i++) ok[arr[i]]++;
    if (ok[arr[n - 1]] > 1) {
      cout << fac[n] << endl;
    } else if (ok[arr[n - 1] - 1] == 0) {
      cout << 0 << endl;
    } else {
      long long minus = 0;
      for (int i = 0; i < n; i++) {
        int left = n - ok[arr[n - 1] - 1] - 1;
        int have = ok[arr[n - 1] - 1];
        if (i < have) continue;
        minus =
            minus + (((nCr(i, have, mod) * fac[have]) % mod) * fac[left]) % mod;
        minus %= mod;
      }
      cout << (fac[n] - minus + mod) % mod << endl;
    }
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}

```