#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 998244353;
const int mxN = 2e5 + 5;
vector<ll> fact(mxN + 5);
ll mul(ll a, ll b) { return (a * b) % MOD; }
ll add(ll a, ll b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
ll sub(ll a, ll b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
ll binpow(ll a, ll n) {
  ll res = 1;
  while (n) {
    if (n & 1) res = mul(res, a);
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}
ll inv(ll x) { return binpow(x, MOD - 2); }
ll C(ll k, ll n) { return mul(fact[n], inv(mul(fact[k], fact[n - k]))); }
void solve() {
  int n;
  cin >> n;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  int mx = *max_element(a.begin(), a.end());
  int cnt = count(a.begin(), a.end(), mx);
  int x = 1, y = 1;
  if (cnt == 1) {
    int maxm = count(a.begin(), a.end(), mx - 1);
    x = maxm + 1;
    y = maxm;
  }
  ll ans = 1;
  for (int i = 1; i <= n; i++) {
    if (i == x)
      ans = mul(ans, y);
    else
      ans = mul(ans, i);
  }
  cout << ans << endl;
}
void cf() {
  int tt;
  cin >> tt;
  while (tt--) solve();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fact[0] = 1;
  for (int i = 1; i <= mxN; i++) fact[i] = mul(fact[i - 1], i);
  cf();
  return 0;
}

### Prompt

Please formulate 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;
using ll = long long;
const ll MOD = 998244353;
const int mxN = 2e5 + 5;
vector<ll> fact(mxN + 5);
ll mul(ll a, ll b) { return (a * b) % MOD; }
ll add(ll a, ll b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
ll sub(ll a, ll b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
ll binpow(ll a, ll n) {
  ll res = 1;
  while (n) {
    if (n & 1) res = mul(res, a);
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}
ll inv(ll x) { return binpow(x, MOD - 2); }
ll C(ll k, ll n) { return mul(fact[n], inv(mul(fact[k], fact[n - k]))); }
void solve() {
  int n;
  cin >> n;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  int mx = *max_element(a.begin(), a.end());
  int cnt = count(a.begin(), a.end(), mx);
  int x = 1, y = 1;
  if (cnt == 1) {
    int maxm = count(a.begin(), a.end(), mx - 1);
    x = maxm + 1;
    y = maxm;
  }
  ll ans = 1;
  for (int i = 1; i <= n; i++) {
    if (i == x)
      ans = mul(ans, y);
    else
      ans = mul(ans, i);
  }
  cout << ans << endl;
}
void cf() {
  int tt;
  cin >> tt;
  while (tt--) solve();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fact[0] = 1;
  for (int i = 1; i <= mxN; i++) fact[i] = mul(fact[i - 1], i);
  cf();
  return 0;
}

```