#include <bits/stdc++.h>
using namespace std;
const int N = 200000 + 9;
const int P = 998244353;
long long fact[N], Inv[N], inv_fact[N];
void find_inv() {
  Inv[0] = 0;
  Inv[1] = 1;
  for (int i = 2; i < N; ++i) Inv[i] = (P - P / i) * Inv[P % i] % P;
  fact[0] = inv_fact[0] = 1;
  for (int i = 1; i < N; ++i) {
    fact[i] = fact[i - 1] * i % P;
    inv_fact[i] = inv_fact[i - 1] * Inv[i] % P;
  }
}
long long C(int n, int m) {
  return fact[n] * inv_fact[m] % P * inv_fact[n - m] % P;
}
void solve() {
  int n;
  scanf("%d", &n);
  map<int, int> mp;
  for (int i = 1; i <= n; ++i) {
    int a;
    scanf("%d", &a);
    ++mp[a];
  }
  if ((int)mp.size() == 1) {
    printf("%lld\n", fact[n]);
  } else if (mp.rbegin()->second >= 2) {
    printf("%lld\n", fact[n]);
  } else if (mp.rbegin()->first - (++mp.rbegin())->first == 1) {
    int x = (++mp.rbegin())->second;
    printf("%lld\n",
           (fact[x + 1] - fact[x] + P) * C(n, x + 1) % P * fact[n - x - 1] % P);
  } else {
    printf("0\n");
  }
}
int main() {
  find_inv();
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

### Prompt

Develop a solution in Cpp 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 int N = 200000 + 9;
const int P = 998244353;
long long fact[N], Inv[N], inv_fact[N];
void find_inv() {
  Inv[0] = 0;
  Inv[1] = 1;
  for (int i = 2; i < N; ++i) Inv[i] = (P - P / i) * Inv[P % i] % P;
  fact[0] = inv_fact[0] = 1;
  for (int i = 1; i < N; ++i) {
    fact[i] = fact[i - 1] * i % P;
    inv_fact[i] = inv_fact[i - 1] * Inv[i] % P;
  }
}
long long C(int n, int m) {
  return fact[n] * inv_fact[m] % P * inv_fact[n - m] % P;
}
void solve() {
  int n;
  scanf("%d", &n);
  map<int, int> mp;
  for (int i = 1; i <= n; ++i) {
    int a;
    scanf("%d", &a);
    ++mp[a];
  }
  if ((int)mp.size() == 1) {
    printf("%lld\n", fact[n]);
  } else if (mp.rbegin()->second >= 2) {
    printf("%lld\n", fact[n]);
  } else if (mp.rbegin()->first - (++mp.rbegin())->first == 1) {
    int x = (++mp.rbegin())->second;
    printf("%lld\n",
           (fact[x + 1] - fact[x] + P) * C(n, x + 1) % P * fact[n - x - 1] % P);
  } else {
    printf("0\n");
  }
}
int main() {
  find_inv();
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

```