#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const int mod = 998244353;
int add(int a, int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
int mul(int a, int b) { return (a * 1ll * b) % mod; }
int sub(int a, int b) {
  if (a - b >= 0)
    return a - b;
  else
    return a - b + mod;
}
int bin_pow(int a, int n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    int b = bin_pow(a, n / 2);
    return mul(b, b);
  } else
    return mul(a, bin_pow(a, n - 1));
}
const int N = 2e5 + 1;
int f[N], rf[N];
int C(int k, int n) { return mul(f[n], mul(rf[k], rf[n - k])); }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  f[0] = f[1] = rf[0] = rf[1] = 1;
  for (int i = 2; i < N; i++) {
    f[i] = mul(f[i - 1], i);
    rf[i] = mul(rf[i - 1], bin_pow(i, mod - 2));
  }
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    map<int, int> cnt;
    for (int i = 0; i < n; i++) {
      int a;
      cin >> a;
      cnt[a]++;
    }
    if (cnt.rbegin()->second >= 2) {
      cout << f[n] << '\n';
    } else {
      auto it = (--(--(cnt.end())));
      if (it->first + 1 != cnt.rbegin()->first) {
        cout << 0 << '\n';
      } else {
        int c = it->second;
        int ans = mul(n - 1, f[n - 1]);
        for (int i = c; i < n - 1; i++) {
          ans = sub(ans, mul(C(c, i), mul(f[c], f[n - 1 - c])));
        }
        cout << ans << '\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>
using namespace std;
const long long inf = 1e18;
const int mod = 998244353;
int add(int a, int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
int mul(int a, int b) { return (a * 1ll * b) % mod; }
int sub(int a, int b) {
  if (a - b >= 0)
    return a - b;
  else
    return a - b + mod;
}
int bin_pow(int a, int n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    int b = bin_pow(a, n / 2);
    return mul(b, b);
  } else
    return mul(a, bin_pow(a, n - 1));
}
const int N = 2e5 + 1;
int f[N], rf[N];
int C(int k, int n) { return mul(f[n], mul(rf[k], rf[n - k])); }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  f[0] = f[1] = rf[0] = rf[1] = 1;
  for (int i = 2; i < N; i++) {
    f[i] = mul(f[i - 1], i);
    rf[i] = mul(rf[i - 1], bin_pow(i, mod - 2));
  }
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    map<int, int> cnt;
    for (int i = 0; i < n; i++) {
      int a;
      cin >> a;
      cnt[a]++;
    }
    if (cnt.rbegin()->second >= 2) {
      cout << f[n] << '\n';
    } else {
      auto it = (--(--(cnt.end())));
      if (it->first + 1 != cnt.rbegin()->first) {
        cout << 0 << '\n';
      } else {
        int c = it->second;
        int ans = mul(n - 1, f[n - 1]);
        for (int i = c; i < n - 1; i++) {
          ans = sub(ans, mul(C(c, i), mul(f[c], f[n - 1 - c])));
        }
        cout << ans << '\n';
      }
    }
  }
  return 0;
}

```