#include <bits/stdc++.h>
using namespace std;
int yh() {
  int ret = 0;
  bool f = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == EOF) return -1;
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) ret = (ret << 3) + (ret << 1) + (c ^ 48), c = getchar();
  return f ? -ret : ret;
}
const int maxn = 3e5 + 5, mod = 998244353;
int n, m;
int a[maxn];
int fac[maxn], ifac[maxn];
int mul(int x, int y) { return (long long)x * y % mod; }
int ksm(int x, int k) {
  int ans = 1;
  for (; k; k >>= 1, x = mul(x, x))
    if (k & 1) ans = mul(ans, x);
  return ans;
}
int C(int x, int y) {
  if (x < y) return 0;
  return mul(fac[x], mul(ifac[y], ifac[x - y]));
}
int main() {
  fac[0] = 1;
  for (int i = (1), iss = (2e5); i <= iss; i++) fac[i] = mul(fac[i - 1], i);
  ifac[200000] = ksm(fac[200000], mod - 2);
  for (int i = (199999), iss = (0); i >= iss; i--)
    ifac[i] = mul(ifac[i + 1], i + 1);
  for (int _ = (yh()), _ss = (1); _ >= _ss; _--) {
    n = yh();
    map<int, int> cnt;
    for (int i = (1), iss = (n); i <= iss; i++) a[i] = yh(), cnt[a[i]]++;
    sort(a + 1, a + 1 + n);
    int ans = fac[n];
    if (cnt[a[n]] > 1) {
      cout << ans << '\n';
      continue;
    }
    if (a[n] - a[n - 1] > 1) {
      cout << 0 << '\n';
      continue;
    }
    int oc = cnt[a[n - 1]];
    cout << (ans - mul(C(n, n - oc - 1), mul(fac[oc], fac[n - oc - 1])) + mod) %
                mod
         << '\n';
  }
  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;
int yh() {
  int ret = 0;
  bool f = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == EOF) return -1;
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) ret = (ret << 3) + (ret << 1) + (c ^ 48), c = getchar();
  return f ? -ret : ret;
}
const int maxn = 3e5 + 5, mod = 998244353;
int n, m;
int a[maxn];
int fac[maxn], ifac[maxn];
int mul(int x, int y) { return (long long)x * y % mod; }
int ksm(int x, int k) {
  int ans = 1;
  for (; k; k >>= 1, x = mul(x, x))
    if (k & 1) ans = mul(ans, x);
  return ans;
}
int C(int x, int y) {
  if (x < y) return 0;
  return mul(fac[x], mul(ifac[y], ifac[x - y]));
}
int main() {
  fac[0] = 1;
  for (int i = (1), iss = (2e5); i <= iss; i++) fac[i] = mul(fac[i - 1], i);
  ifac[200000] = ksm(fac[200000], mod - 2);
  for (int i = (199999), iss = (0); i >= iss; i--)
    ifac[i] = mul(ifac[i + 1], i + 1);
  for (int _ = (yh()), _ss = (1); _ >= _ss; _--) {
    n = yh();
    map<int, int> cnt;
    for (int i = (1), iss = (n); i <= iss; i++) a[i] = yh(), cnt[a[i]]++;
    sort(a + 1, a + 1 + n);
    int ans = fac[n];
    if (cnt[a[n]] > 1) {
      cout << ans << '\n';
      continue;
    }
    if (a[n] - a[n - 1] > 1) {
      cout << 0 << '\n';
      continue;
    }
    int oc = cnt[a[n - 1]];
    cout << (ans - mul(C(n, n - oc - 1), mul(fac[oc], fac[n - oc - 1])) + mod) %
                mod
         << '\n';
  }
  return 0;
}

```