description: 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.
solution:  //package eround101;
  
  import java.util.*;
  import java.io.*;
  import java.lang.*;
  import java.util.StringTokenizer;
  
  public class C101 {
  
      static HritikScanner sc = new HritikScanner();
      static PrintWriter pw = new PrintWriter(System.out, true);
      final static int MOD = (int) 1e9 + 7;
      static StringBuilder sb = new StringBuilder();
  
      public static void main(String[] args) {
  
          int t = ni();
          while (t-- > 0) {
              solve();
          }
      }
  
      static void solve() {
          int n = ni();
          int[] arr = nextIntArray(n);
          int c = 0;
          int i;
          for(i = 1; i <= n; i++)
          {
          	int j = 0;
          	if(i % 2 == 0)
          		j = 1;
                  if(isSorted(arr))
                  {
                      pl(i-1);
                      return;
                  }
          	for(; j < n-1; j = j+2)
          	{
          		if(arr[j] > arr[j+1])
          		{
          			int temp = arr[j];
          			arr[j] = arr[j+1];
          			arr[j+1] = temp;
          		}
          	}
          }
          System.out.println(i-1);
      
      }
      static boolean isSorted(int[] arr)
      {
          for(int i = 0; i < arr.length - 1; i++)
          {
              if(arr[i] > arr[i+1])
                  return false;
          }
          return true;
      }
      
      
      
      
      /////////////////////////////////////////////////////////////////////////////////
      static int[] nextIntArray(int n) {
          int[] arr = new int[n];
          int i = 0;
          while (i < n) {
              arr[i++] = ni();
          }
          return arr;
      }
  
      static long[] nextLongArray(int n) {
          long[] arr = new long[n];
          int i = 0;
          while (i < n) {
              arr[i++] = nl();
          }
          return arr;
      }
  
      static int[] nextIntArray1(int n) {
          int[] arr = new int[n + 1];
          int i = 1;
          while (i <= n) {
              arr[i++] = ni();
          }
          return arr;
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      static int ni() {
          return sc.nextInt();
      }
  
      static long nl() {
          return sc.nextLong();
      }
  
      static double nd() {
          return sc.nextDouble();
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      static void pl() {
          pw.println();
      }
  
      static void p(Object o) {
          pw.print(o+ " ");
      }
  
      static void pl(Object o) {
          pw.println(o);
      }
  
      static void psb(StringBuilder sb) {
          pw.print(sb);
      }
  
      static void pa(Object arr[]) {
          for (Object o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(int arr[]) {
          for (int o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(long arr[]) {
          for (long o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(double arr[]) {
          for (double o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(char arr[]) {
          for (char o : arr) {
              p(o);
          }
          pl();
      }
  
      static void pa(List list) {
          for (Object o : list) {
              p(o);
          }
          pl();
      }
  
      static void pa(Object[][] arr) {
  
          for (int i = 0; i < arr.length; ++i) {
              for (Object o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(int[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (int o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(long[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (long o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(char[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (char o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
      static void pa(double[][] arr) {
          for (int i = 0; i < arr.length; ++i) {
              for (double o : arr[i]) {
                  p(o);
              }
              pl();
          }
      }
  
  /////////////////////////////////////////////////////////////////////////////////
      static void print(Object s) {
          System.out.println(s);
      }
  
      /////////////////////////////////////////////////////////////////////////////////
      //-----------HritikScanner class for faster input----------//
      static class HritikScanner {
  
          BufferedReader br;
          StringTokenizer st;
  
          public HritikScanner() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      //////////////////////////////////////////////////////////////////
      public static class Pair implements Comparable<Pair> {
  
          int num, row,col;
  
          Pair(int num, int row, int col) {
              this.num = num;
              this.row = row;
              this.col = col;
          }
  
          public int get1() {
              return num;
          }
  
          public int get2() {
              return row;
          }
  
          public int compareTo(Pair A) {
              return this.num - A.num;
          }
          
          public String toString()
          {
              return num +" "+ row+" "+col;
          }
      }
  
      //////////////////////////////////////////////////////////////////
      // Function to return gcd of a and b time complexity O(log(a+b))
      static long gcd(long a, long b) {
          if (a == 0) {
              return b;
          }
          return gcd(b % a, a);
      }
  
      // method to return LCM of two numbers 
      static long lcm(int a, int b) {
          return (a / gcd(a, b)) * b;
      }
  
      //////////////////////////////////////////////////////////////////
      static boolean isPrime(long n) {
          // Corner cases 
          if (n <= 1) {
              return false;
          }
          if (n <= 3) {
              return true;
          }
  
          // This is checked so that we can skip  
          // middle five numbers in below loop 
          if (n % 2 == 0 || n % 3 == 0) {
              return false;
          }
  
          for (long i = 5; i * i <= n; i = i + 6) {
              if (n % i == 0 || n % (i + 2) == 0) {
                  return false;
              }
          }
  
          return true;
      }
      //////////////////////////////////////////////////////////////////
  
      static boolean isPowerOfTwo(long n) {
          if (n == 0) {
              return false;
          }
  
          return (long) (Math.ceil((Math.log(n) / Math.log(2))))
                  == (long) (Math.floor(((Math.log(n) / Math.log(2)))));
      }
  
      public static long getFact(int n) {
          long ans = 1;
          while (n > 0) {
              ans *= n;
              ans %= MOD;
              n--;
          }
          return ans;
      }
  
      public static long pow(long n, int pow) {
          if (pow == 0) {
              return 1;
          }
          long temp = pow(n, pow / 2) % MOD;
          temp *= temp;
          temp %= MOD;
          if (pow % 2 == 1) {
              temp *= n;
          }
          temp %= MOD;
          return temp;
  
      }
  
      public static long nCr(int n, int r) {
          long ans = 1;
          int temp = n - r;
          while (n > temp) {
              ans *= n;
              ans %= MOD;
              n--;
          }
          ans *= pow(getFact(r) % MOD, MOD - 2) % MOD;
          ans %= MOD;
          return ans;
      }
  
      //////////////////////////////////////////////////////////////////
      // method returns Nth power of A 
      static double nthRoot(int A, int N) {
  
          // intially guessing a random number between 
          // 0 and 9 
          double xPre = Math.random() % 10;
  
          // smaller eps, denotes more accuracy 
          double eps = 0.001;
  
          // initializing difference between two 
          // roots by INT_MAX 
          double delX = 2147483647;
  
          // xK denotes current value of x 
          double xK = 0.0;
  
          // loop untill we reach desired accuracy 
          while (delX > eps) {
              // calculating current value from previous 
              // value by newton's method 
              xK = ((N - 1.0) * xPre
                      + (double) A / Math.pow(xPre, N - 1)) / (double) N;
              delX = Math.abs(xK - xPre);
              xPre = xK;
          }
  
          return xK;
      }
  
  }
  