#include <bits/stdc++.h>
using namespace std;
vector<long long> fact(2000010);
vector<long long> inv(2000010);
const long long mod = 998244353;
long long modpow(long long x, long long y) {
  return !y ? 1 : ((y % 2 ? x : 1) * modpow((x * x) % mod, y / 2)) % mod;
}
long long choose(long long x, long long y) {
  if (y > x) return 0;
  return (fact[x] * inv[y] % mod) * inv[x - y] % mod;
}
long long permute(long long x, long long y) {
  if (y > x) return 0;
  return (fact[x] * inv[x - y]) % mod;
}
void precalc() {
  fact[0] = 1;
  inv[0] = 1;
  for (long long i = 1; i <= 2000000; i++) {
    fact[i] = fact[i - 1] * i % mod;
  }
  inv[2000000] = modpow(fact[2000000], mod - 2);
  for (long long i = 1999999; i > 0; i--) {
    inv[i] = inv[i + 1] * (i + 1) % mod;
  }
}
signed main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  precalc();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> arr(n);
    map<long long, long long> freq;
    set<long long> s;
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
      s.insert(arr[i]);
      freq[arr[i]]++;
    }
    sort(arr.begin(), arr.end());
    if (arr[n - 2] == arr[n - 1]) {
      cout << fact[n] << "\n";
    } else if (arr[n - 2] + 1 == arr[n - 1]) {
      long long total = fact[n];
      long long rem = n - freq[arr[n - 1]] - freq[arr[n - 2]];
      long long val = fact[rem];
      long long cur = 0;
      for (long long i = 1; i <= n; i++) {
        cur += permute(i - 1, freq[arr[n - 2]]);
        cur %= mod;
      }
      total -= val * cur % mod;
      total += mod;
      total %= mod;
      cout << total << "\n";
    } else {
      cout << "0\n";
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
vector<long long> fact(2000010);
vector<long long> inv(2000010);
const long long mod = 998244353;
long long modpow(long long x, long long y) {
  return !y ? 1 : ((y % 2 ? x : 1) * modpow((x * x) % mod, y / 2)) % mod;
}
long long choose(long long x, long long y) {
  if (y > x) return 0;
  return (fact[x] * inv[y] % mod) * inv[x - y] % mod;
}
long long permute(long long x, long long y) {
  if (y > x) return 0;
  return (fact[x] * inv[x - y]) % mod;
}
void precalc() {
  fact[0] = 1;
  inv[0] = 1;
  for (long long i = 1; i <= 2000000; i++) {
    fact[i] = fact[i - 1] * i % mod;
  }
  inv[2000000] = modpow(fact[2000000], mod - 2);
  for (long long i = 1999999; i > 0; i--) {
    inv[i] = inv[i + 1] * (i + 1) % mod;
  }
}
signed main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  precalc();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> arr(n);
    map<long long, long long> freq;
    set<long long> s;
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
      s.insert(arr[i]);
      freq[arr[i]]++;
    }
    sort(arr.begin(), arr.end());
    if (arr[n - 2] == arr[n - 1]) {
      cout << fact[n] << "\n";
    } else if (arr[n - 2] + 1 == arr[n - 1]) {
      long long total = fact[n];
      long long rem = n - freq[arr[n - 1]] - freq[arr[n - 2]];
      long long val = fact[rem];
      long long cur = 0;
      for (long long i = 1; i <= n; i++) {
        cur += permute(i - 1, freq[arr[n - 2]]);
        cur %= mod;
      }
      total -= val * cur % mod;
      total += mod;
      total %= mod;
      cout << total << "\n";
    } else {
      cout << "0\n";
    }
  }
  return 0;
}

```