Educational Codeforces Round 113 (Rated for Div. 2)C. Jury Meeting

C. Jury Meeting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

nn people gathered to hold a jury meeting of the upcoming competition, the ii-th member of the jury came up with aiai 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 pp of numbers from 11 to nn (an array of size nn where each integer from 11 to nn occurs exactly once).

Then the discussion goes as follows:

• If a jury member p1p1 has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped.
• If a jury member p2p2 has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped.
• ...
• If a jury member pnpn 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 pp 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 998244353998244353.

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The first line of the test case contains a single integer nn (2≤n≤2⋅1052≤n≤2⋅105) — number of jury members.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the number of problems that the ii-th member of the jury came up with.

The sum of nn over all test cases does not exceed 2⋅1052⋅105.

Output

For each test case, print one integer — the number of nice permutations, taken modulo 998244353998244353.

Example

input

Copy

4

2

1 2

3

5 5 5

4

1 3 3 7

6

3 4 2 1 3 3

output

Copy

1

6

0

540

Note

Explanation of the first test case from the example:

There are two possible permutations, p=[1,2]p=[1,2] and p=[2,1]p=[2,1]. For p=[1,2]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]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.

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 2e5 + 5;
const int M = 1e5 + 5;
const ll mod = 998244353;
ll qpow(ll a, ll b) {//快速幂改编用法
  ll res = 1;
  while (b) {
    if (b & 1)
      res = res * a % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
int n, m, t;
ll a[N];
ll pre[N];//pre[i]=1*2*...*i
ll A(ll nn, ll mm) {
  return pre[nn] * qpow(pre[nn - mm], mod - 2) % mod;
}
void solve() {
  cin >> n;
  ll maxx = -1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  if (a[n - 1] >= a[n - 2] + 2)//发现bug，爬5 3 2>>0
    cout << 0 << '\n';
  else if (a[n - 1] == a[n - 2])//发现新世界，起飞 5 5 5>>6
    cout << pre[n] << '\n';
  else {//出现狗屎一样的第三种情况 5 4 3 >>3
    int cnt = 0;
    for (int i = 0; i < n; i++) {//并列第二名有几个
      if (a[i] == a[n - 2])
        cnt++;//最大值不能在所有次大值的最右边
    }
    ll ans = pre[n];//全排列>>意味着所有情况
    for (int i = n; i >= cnt + 1; i--) { //i是最大值的位置 枚举i所在位置不是好序列的方案数
      ll res = (A(i - 1, cnt) * pre[n - cnt - 1]) % mod;//此处难点
      //cout<<res<<'\n';
      ans = (ans - res + mod) % mod;
    }
    cout << ans % mod << '\n';
  }
}
int main() {
  cin >> t;
  pre[0] = 1;
  for (int i = 1; i <= 2e5; i++) {
    pre[i] = (pre[i - 1] * i) % mod;//全排列
  }
  while (t--) {
    solve();
  }
}