#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long int mod = 998244353;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long int inv(long long int a) { return power(a, mod - 2); }
long long int add(long long int a, long long int b) { return (a + b) % mod; }
long long int sub(long long int a, long long int b) {
  return (a - b + mod) % mod;
}
long long int mul(long long int a, long long int b) { return (a * b) % mod; }
long long int divi(long long int a, long long int b) {
  return (a * inv(b)) % mod;
}
long long fact[300000 + 1];
void pre() {
  fact[0] = 1;
  for (int i = 1; i < 300000; i++) fact[i] = mul(fact[i - 1], i);
}
void solve() {
  int n;
  cin >> n;
  long long arr[n];
  map<long long, long long> m;
  long long ma = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    m[arr[i]]++;
    ma = max(arr[i], ma);
  }
  if (m[ma] > 1)
    cout << fact[n] << "\n";
  else
    cout << sub(fact[n], divi(fact[n], m[ma] + m[ma - 1])) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre();
  long long int num = 1;
  cin >> num;
  for (long long int i = 0; i < num; i++) {
    solve();
  }
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long int mod = 998244353;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long int inv(long long int a) { return power(a, mod - 2); }
long long int add(long long int a, long long int b) { return (a + b) % mod; }
long long int sub(long long int a, long long int b) {
  return (a - b + mod) % mod;
}
long long int mul(long long int a, long long int b) { return (a * b) % mod; }
long long int divi(long long int a, long long int b) {
  return (a * inv(b)) % mod;
}
long long fact[300000 + 1];
void pre() {
  fact[0] = 1;
  for (int i = 1; i < 300000; i++) fact[i] = mul(fact[i - 1], i);
}
void solve() {
  int n;
  cin >> n;
  long long arr[n];
  map<long long, long long> m;
  long long ma = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    m[arr[i]]++;
    ma = max(arr[i], ma);
  }
  if (m[ma] > 1)
    cout << fact[n] << "\n";
  else
    cout << sub(fact[n], divi(fact[n], m[ma] + m[ma - 1])) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre();
  long long int num = 1;
  cin >> num;
  for (long long int i = 0; i < num; i++) {
    solve();
  }
}

```