#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
void chkmax(long long &a, long long b) {
  if (a < b) a = b;
}
void chkmin(long long &a, long long b) {
  if (a > b) a = b;
}
long long qpow(long long d, long long z) {
  long long res = 1;
  for (; z; z >>= 1) {
    if (z & 1) res = res * d % MOD;
    d = d * d % MOD;
  }
  return res;
}
int n;
int d[200005];
void solve() {
  scanf("%d", &n);
  int Max = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &d[i]);
    if (Max < d[i]) Max = d[i];
  }
  int cM0 = 0, cM1 = 0;
  for (int i = 1; i <= n; i++) {
    if (d[i] == Max)
      cM0++;
    else if (d[i] + 1 == Max)
      cM1++;
  }
  long long ff = 1;
  for (int i = 1; i <= n; i++) ff = ff * i % MOD;
  assert(cM0 >= 1);
  if (cM0 >= 2) {
    long long ans = 1;
    for (int i = 1; i <= n; i++) ans = ans * i % MOD;
    printf("%lld\n", ans);
  } else if (cM1 == 0) {
    printf("0\n");
  } else {
    long long ans = 1;
    for (int i = 0; i < cM1 + 1; i++) ans = ans * (n - i) % MOD;
    for (int i = 1; i <= n - cM1 - 1; i++) ans = ans * i % MOD;
    ans = ans * qpow(cM1 + 1, MOD - 2) % MOD;
    ans = (ff - ans + MOD) % MOD;
    printf("%lld\n", ans);
  }
  return;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long MOD = 998244353;
void chkmax(long long &a, long long b) {
  if (a < b) a = b;
}
void chkmin(long long &a, long long b) {
  if (a > b) a = b;
}
long long qpow(long long d, long long z) {
  long long res = 1;
  for (; z; z >>= 1) {
    if (z & 1) res = res * d % MOD;
    d = d * d % MOD;
  }
  return res;
}
int n;
int d[200005];
void solve() {
  scanf("%d", &n);
  int Max = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &d[i]);
    if (Max < d[i]) Max = d[i];
  }
  int cM0 = 0, cM1 = 0;
  for (int i = 1; i <= n; i++) {
    if (d[i] == Max)
      cM0++;
    else if (d[i] + 1 == Max)
      cM1++;
  }
  long long ff = 1;
  for (int i = 1; i <= n; i++) ff = ff * i % MOD;
  assert(cM0 >= 1);
  if (cM0 >= 2) {
    long long ans = 1;
    for (int i = 1; i <= n; i++) ans = ans * i % MOD;
    printf("%lld\n", ans);
  } else if (cM1 == 0) {
    printf("0\n");
  } else {
    long long ans = 1;
    for (int i = 0; i < cM1 + 1; i++) ans = ans * (n - i) % MOD;
    for (int i = 1; i <= n - cM1 - 1; i++) ans = ans * i % MOD;
    ans = ans * qpow(cM1 + 1, MOD - 2) % MOD;
    ans = (ff - ans + MOD) % MOD;
    printf("%lld\n", ans);
  }
  return;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}

```