#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 10;
long long mod = 998244353;
long long a[maxn];
long long fac[maxn];
map<long long, long long> mp;
long long qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
long long c(long long m, long long n) {
  return fac[m] * qpow(fac[n] * fac[m - n] % mod, mod - 2) % mod;
}
int main() {
  int t;
  cin >> t;
  fac[0] = 1;
  for (int i = 1; i < maxn; i++) fac[i] = fac[i - 1] * i % mod;
  while (t--) {
    int n;
    mp.clear();
    cin >> n;
    long long ma = 0;
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      scanf("%lld", &a[i]), mp[a[i]]++, ma = max(ma, a[i]);
    int sum = 0;
    if (mp[ma] >= 2)
      printf("%lld\n", fac[n]);
    else if (mp[ma] == 1) {
      if (mp[ma - 1] == 0) {
        printf("0\n");
      } else {
        long long sum = 0;
        for (int i = mp[ma - 1]; i <= n - 1; i++) {
          sum += fac[i] * fac[n - i - 1] % mod *
                 c(n - mp[ma - 1] - 1, i - mp[ma - 1]) % mod;
          sum %= mod;
        }
        long long ans = (fac[n] - sum) % mod + mod;
        printf("%lld\n", ans % mod);
      }
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 int maxn = 5e5 + 10;
long long mod = 998244353;
long long a[maxn];
long long fac[maxn];
map<long long, long long> mp;
long long qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
long long c(long long m, long long n) {
  return fac[m] * qpow(fac[n] * fac[m - n] % mod, mod - 2) % mod;
}
int main() {
  int t;
  cin >> t;
  fac[0] = 1;
  for (int i = 1; i < maxn; i++) fac[i] = fac[i - 1] * i % mod;
  while (t--) {
    int n;
    mp.clear();
    cin >> n;
    long long ma = 0;
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      scanf("%lld", &a[i]), mp[a[i]]++, ma = max(ma, a[i]);
    int sum = 0;
    if (mp[ma] >= 2)
      printf("%lld\n", fac[n]);
    else if (mp[ma] == 1) {
      if (mp[ma - 1] == 0) {
        printf("0\n");
      } else {
        long long sum = 0;
        for (int i = mp[ma - 1]; i <= n - 1; i++) {
          sum += fac[i] * fac[n - i - 1] % mod *
                 c(n - mp[ma - 1] - 1, i - mp[ma - 1]) % mod;
          sum %= mod;
        }
        long long ans = (fac[n] - sum) % mod + mod;
        printf("%lld\n", ans % mod);
      }
    }
  }
  return 0;
}

```