#include <bits/stdc++.h>
using namespace std;
long long i, j, k, n, m, M, s, len, ans, co, T, f, r, t, pos, pre, now;
long long mod = 998244353, mmax, mmax2, co1, co2;
long long fact[200001], inv[200001], invfac[200001];
long long a[200001];
unordered_map<long long, long long> z;
void init(long long n = 200000) {
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) fact[i] = (long long)i * fact[i - 1] % mod;
  inv[1] = 1;
  for (long long i = 2; i <= n; i++)
    inv[i] = mod - (long long)(mod / i) * inv[mod % i] % mod;
  invfac[0] = 1;
  for (long long i = 1; i <= n; i++)
    invfac[i] = (long long)invfac[i - 1] * inv[i] % mod;
}
long long C(long long n, long long m) {
  return (long long)fact[n] * invfac[m] % mod * invfac[n - m] % mod;
}
int main() {
  init();
  scanf("%lld", &T);
  while (T--) {
    z.clear();
    mmax = mmax2 = -1;
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      z[a[i]]++;
    }
    for (i = 1; i <= n; i++) mmax = max(mmax, a[i]);
    for (i = 1; i <= n; i++)
      if (a[i] != mmax) mmax2 = max(mmax2, a[i]);
    co1 = z[mmax], co2 = z[mmax2];
    if (mmax2 == -1 || z[mmax] >= 2)
      ans = fact[n] % mod;
    else if (mmax - mmax2 >= 2)
      ans = 0;
    else
      ans = C(n, co1 + co2) * C(co2, 1) % mod * fact[co2 + co1 - 1] % mod *
            fact[n - co1 - co2] % mod;
    printf("%lld\n", ans);
  }
  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;
long long i, j, k, n, m, M, s, len, ans, co, T, f, r, t, pos, pre, now;
long long mod = 998244353, mmax, mmax2, co1, co2;
long long fact[200001], inv[200001], invfac[200001];
long long a[200001];
unordered_map<long long, long long> z;
void init(long long n = 200000) {
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) fact[i] = (long long)i * fact[i - 1] % mod;
  inv[1] = 1;
  for (long long i = 2; i <= n; i++)
    inv[i] = mod - (long long)(mod / i) * inv[mod % i] % mod;
  invfac[0] = 1;
  for (long long i = 1; i <= n; i++)
    invfac[i] = (long long)invfac[i - 1] * inv[i] % mod;
}
long long C(long long n, long long m) {
  return (long long)fact[n] * invfac[m] % mod * invfac[n - m] % mod;
}
int main() {
  init();
  scanf("%lld", &T);
  while (T--) {
    z.clear();
    mmax = mmax2 = -1;
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      z[a[i]]++;
    }
    for (i = 1; i <= n; i++) mmax = max(mmax, a[i]);
    for (i = 1; i <= n; i++)
      if (a[i] != mmax) mmax2 = max(mmax2, a[i]);
    co1 = z[mmax], co2 = z[mmax2];
    if (mmax2 == -1 || z[mmax] >= 2)
      ans = fact[n] % mod;
    else if (mmax - mmax2 >= 2)
      ans = 0;
    else
      ans = C(n, co1 + co2) * C(co2, 1) % mod * fact[co2 + co1 - 1] % mod *
            fact[n - co1 - co2] % mod;
    printf("%lld\n", ans);
  }
  return 0;
}

```