#include <bits/stdc++.h>
using namespace std;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
const int64_t M = 998244353;
const long long MAX = 200001;
int64_t power(int64_t a, int64_t b) {
  b = (b < 0) ? (M + b - 1) : b;
  int64_t res = 1LL;
  while (b) {
    if (b % 2) res = res * a % M;
    a = a * a % M;
    b /= 2;
  }
  return res;
}
vector<long long> fac;
void prepareFactorial() {
  fac.resize(MAX);
  fac[0] = fac[1] = 1;
  for (long long i = 2; i < MAX; i++) fac[i] = (fac[i - 1] * i) % M;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> jury(n);
  unordered_map<long long, long long, custom_hash> frequency;
  pair<long long, long long> max_pair;
  for (long long &task : jury) {
    cin >> task;
    frequency[task]++;
    if (task > max_pair.first) {
      swap(max_pair.first, max_pair.second);
      max_pair.first = task;
    } else if (task > max_pair.second) {
      max_pair.second = task;
    }
  }
  long long cntMx = frequency[max_pair.first];
  long long cntSecondMx = frequency[max_pair.second];
  long long ans = 0;
  if (cntMx > 1) {
    ans = fac[n];
  } else if (max_pair.first - max_pair.second == 1) {
    long long total = fac[n];
    long long invalid = (fac[n] * power(cntSecondMx + 1, -1)) % M;
    ans = (total - invalid + M) % M;
  }
  cout << ans << "\n";
}
int32_t main() {
  prepareFactorial();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Create a solution in cpp for 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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
const int64_t M = 998244353;
const long long MAX = 200001;
int64_t power(int64_t a, int64_t b) {
  b = (b < 0) ? (M + b - 1) : b;
  int64_t res = 1LL;
  while (b) {
    if (b % 2) res = res * a % M;
    a = a * a % M;
    b /= 2;
  }
  return res;
}
vector<long long> fac;
void prepareFactorial() {
  fac.resize(MAX);
  fac[0] = fac[1] = 1;
  for (long long i = 2; i < MAX; i++) fac[i] = (fac[i - 1] * i) % M;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> jury(n);
  unordered_map<long long, long long, custom_hash> frequency;
  pair<long long, long long> max_pair;
  for (long long &task : jury) {
    cin >> task;
    frequency[task]++;
    if (task > max_pair.first) {
      swap(max_pair.first, max_pair.second);
      max_pair.first = task;
    } else if (task > max_pair.second) {
      max_pair.second = task;
    }
  }
  long long cntMx = frequency[max_pair.first];
  long long cntSecondMx = frequency[max_pair.second];
  long long ans = 0;
  if (cntMx > 1) {
    ans = fac[n];
  } else if (max_pair.first - max_pair.second == 1) {
    long long total = fac[n];
    long long invalid = (fac[n] * power(cntSecondMx + 1, -1)) % M;
    ans = (total - invalid + M) % M;
  }
  cout << ans << "\n";
}
int32_t main() {
  prepareFactorial();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```