#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pll = pair<ll, ll>;
using t4 = tuple<ll, ll, ll, ll>;
int T;
const int mod = 998244353;
ll fact[220000];
ll invfact[220000];
ll modpow(ll a, ll k) {
  if (k == 0) return 1;
  if (k & 1)
    return modpow(a, k - 1) * a % mod;
  else {
    ll half = modpow(a, k / 2);
    return half * half % mod;
  }
}
ll nCk(ll n, ll k) {
  if (k == 0 || n == k) return 1;
  return fact[n] * invfact[k] % mod * invfact[n - k] % mod;
}
void f() {
  int N;
  cin >> N;
  vector<ll> v(N);
  for (int i = 0; i < N; i++) cin >> v[i];
  ll mx = *max_element(v.begin(), v.end());
  int cnt = 0;
  for (int i = 0; i < N; i++)
    if (mx == v[i]) cnt++;
  if (cnt >= 2) {
    cout << fact[N] << '\n';
  } else {
    int mx2 = 0, cnt2 = 0;
    for (int i = 0; i < N; i++) {
      if (v[i] < mx && v[i] > mx2) {
        mx2 = v[i];
        cnt2 = 1;
      } else if (mx2 == v[i])
        cnt2++;
    }
    if (mx2 != mx - 1) {
      cout << 0 << '\n';
      return;
    }
    ll ans = fact[N];
    ll tmp = fact[cnt2] * fact[N - cnt2 - 1] % mod;
    tmp = tmp * nCk(N, N - cnt2 - 1) % mod;
    ans = (ans - tmp + mod) % mod;
    cout << ans << '\n';
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> T;
  fact[0] = 1;
  for (int i = 1; i <= 200000; i++) fact[i] = fact[i - 1] * i % mod;
  invfact[200000] = modpow(fact[200000], mod - 2);
  for (int i = 200000; i >= 1; i--) invfact[i - 1] = invfact[i] * i % mod;
  while (T--) f();
  return 0;
}

### Prompt

Generate 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;
using pll = pair<ll, ll>;
using t4 = tuple<ll, ll, ll, ll>;
int T;
const int mod = 998244353;
ll fact[220000];
ll invfact[220000];
ll modpow(ll a, ll k) {
  if (k == 0) return 1;
  if (k & 1)
    return modpow(a, k - 1) * a % mod;
  else {
    ll half = modpow(a, k / 2);
    return half * half % mod;
  }
}
ll nCk(ll n, ll k) {
  if (k == 0 || n == k) return 1;
  return fact[n] * invfact[k] % mod * invfact[n - k] % mod;
}
void f() {
  int N;
  cin >> N;
  vector<ll> v(N);
  for (int i = 0; i < N; i++) cin >> v[i];
  ll mx = *max_element(v.begin(), v.end());
  int cnt = 0;
  for (int i = 0; i < N; i++)
    if (mx == v[i]) cnt++;
  if (cnt >= 2) {
    cout << fact[N] << '\n';
  } else {
    int mx2 = 0, cnt2 = 0;
    for (int i = 0; i < N; i++) {
      if (v[i] < mx && v[i] > mx2) {
        mx2 = v[i];
        cnt2 = 1;
      } else if (mx2 == v[i])
        cnt2++;
    }
    if (mx2 != mx - 1) {
      cout << 0 << '\n';
      return;
    }
    ll ans = fact[N];
    ll tmp = fact[cnt2] * fact[N - cnt2 - 1] % mod;
    tmp = tmp * nCk(N, N - cnt2 - 1) % mod;
    ans = (ans - tmp + mod) % mod;
    cout << ans << '\n';
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> T;
  fact[0] = 1;
  for (int i = 1; i <= 200000; i++) fact[i] = fact[i - 1] * i % mod;
  invfact[200000] = modpow(fact[200000], mod - 2);
  for (int i = 200000; i >= 1; i--) invfact[i - 1] = invfact[i] * i % mod;
  while (T--) f();
  return 0;
}

```