#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
  ll n;
  cin >> n;
  vector<ll> a(n);
  for (ll i = 0; i < n; i++) cin >> a[i];
  vector<ll> b = a;
  sort(b.begin(), b.end());
  for (ll i = 0; i < n; i++) {
    if (a == b) {
      cout << i << endl;
      return;
    }
    if (i % 2 == 0) {
      ll j = 0;
      ll t = n / 2;
      while (t--) {
        if (a[j] > a[j + 1]) swap(a[j], a[j + 1]);
        j = j + 2;
      }
    }
    if (i % 2 == 1) {
      ll j = 1;
      ll t = n / 2;
      while (t--) {
        if (a[j] > a[j + 1]) swap(a[j], a[j + 1]);
        j = j + 2;
      }
    }
  }
  cout << n << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please formulate a cpp solution to the following problem:

You have a permutation: an array a = [a_1, a_2, …, a_n] of distinct integers from 1 to n. The length of the permutation n is odd.

Consider the following algorithm of sorting the permutation in increasing order.

A helper procedure of the algorithm, f(i), takes a single argument i (1 ≤ i ≤ n-1) and does the following. If a_i > a_{i+1}, the values of a_i and a_{i+1} are exchanged. Otherwise, the permutation doesn't change.

The algorithm consists of iterations, numbered with consecutive integers starting with 1. On the i-th iteration, the algorithm does the following: 

  * if i is odd, call f(1), f(3), …, f(n - 2); 
  * if i is even, call f(2), f(4), …, f(n - 1). 



It can be proven that after a finite number of iterations the permutation will be sorted in increasing order.

After how many iterations will this happen for the first time?

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first line of each test case contains a single integer n (3 ≤ n ≤ 999; n is odd) — the length of the permutation.

The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — the permutation itself. 

It is guaranteed that the sum of n over all test cases does not exceed 999.

Output

For each test case print the number of iterations after which the permutation will become sorted in increasing order for the first time.

If the given permutation is already sorted, print 0.

Example

Input


3
3
3 2 1
7
4 5 7 1 3 2 6
5
1 2 3 4 5


Output


3
5
0

Note

In the first test case, the permutation will be changing as follows: 

  * after the 1-st iteration: [2, 3, 1]; 
  * after the 2-nd iteration: [2, 1, 3]; 
  * after the 3-rd iteration: [1, 2, 3]. 



In the second test case, the permutation will be changing as follows: 

  * after the 1-st iteration: [4, 5, 1, 7, 2, 3, 6]; 
  * after the 2-nd iteration: [4, 1, 5, 2, 7, 3, 6]; 
  * after the 3-rd iteration: [1, 4, 2, 5, 3, 7, 6]; 
  * after the 4-th iteration: [1, 2, 4, 3, 5, 6, 7]; 
  * after the 5-th iteration: [1, 2, 3, 4, 5, 6, 7]. 



In the third test case, the permutation is already sorted and the answer is 0.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
  ll n;
  cin >> n;
  vector<ll> a(n);
  for (ll i = 0; i < n; i++) cin >> a[i];
  vector<ll> b = a;
  sort(b.begin(), b.end());
  for (ll i = 0; i < n; i++) {
    if (a == b) {
      cout << i << endl;
      return;
    }
    if (i % 2 == 0) {
      ll j = 0;
      ll t = n / 2;
      while (t--) {
        if (a[j] > a[j + 1]) swap(a[j], a[j + 1]);
        j = j + 2;
      }
    }
    if (i % 2 == 1) {
      ll j = 1;
      ll t = n / 2;
      while (t--) {
        if (a[j] > a[j + 1]) swap(a[j], a[j + 1]);
        j = j + 2;
      }
    }
  }
  cout << n << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```