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: import java.io.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class Main {
      public static void main(String[] args){
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          InputReader in = new InputReader(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          solve(in, out);
          out.close();
      }
  
      static String reverse(String s) {
          return (new StringBuilder(s)).reverse().toString();
      }
  
      static void sieveOfEratosthenes(int n, int factors[], ArrayList<Integer> ar) 
      { 
          factors[1]=1;
          int p;
          for(p = 2; p*p <=n; p++) 
          { 
              if(factors[p] == 0) 
              { 
                  ar.add(p);
                  factors[p]=p;
                  for(int i = p*p; i <= n; i += p) 
                      if(factors[i]==0)
                          factors[i] = p; 
              } 
          } 
          for(;p<=n;p++){
              if(factors[p] == 0) 
              { 
                  factors[p] = p;
                  ar.add(p);
              } 
          }
      }
  
      static void sort(int ar[]) {
          int n = ar.length;
          ArrayList<Integer> a = new ArrayList<>();
          for (int i = 0; i < n; i++)
              a.add(ar[i]);
          Collections.sort(a);
          for (int i = 0; i < n; i++)
              ar[i] = a.get(i);
      }
  
      static void sort1(long ar[]) {
          int n = ar.length;
          ArrayList<Long> a = new ArrayList<>();
          for (int i = 0; i < n; i++)
              a.add(ar[i]);
          Collections.sort(a);
          for (int i = 0; i < n; i++)
              ar[i] = a.get(i);
      }
  
      static long ncr(long n, long r, long mod) {
          if (r == 0)
              return 1;
          long val = ncr(n - 1, r - 1, mod);
          val = (n * val) % mod;
          val = (val * modInverse(r, mod)) % mod;
          return val;
      }
  
      static int findMax(int a[], int n, int vis[], int i, int d){
          if(i>=n)
              return 0;
          if(vis[i]==1)
              return findMax(a, n, vis, i+1, d);
          int max = 0;
          for(int j=i+1;j<n;j++){
              if(Math.abs(a[i]-a[j])>d||vis[j]==1)
                  continue;
              vis[j] = 1;
              max = Math.max(max, 1 + findMax(a, n, vis, i+1, d));
              vis[j] = 0;
          }
          return max;
      }
  
      static void findSub(ArrayList<ArrayList<Integer>> ar, int n, ArrayList<Integer> a, int i){
          if(i==n){
              ArrayList<Integer> b = new ArrayList<Integer>();
              for(int y:a){
                  b.add(y);
              }
              ar.add(b);
              return;
          }
          for(int j=0;j<n;j++){
              if(j==i)
                  continue;
              a.set(i,j);
              findSub(ar, n, a, i+1);
          }
      }
      
      public static void solve(InputReader sc, PrintWriter pw){
          int t=sc.nextInt();
          // int t=1;
          while(--t>=0){
              int n=sc.nextInt();
              int a[]=sc.readArray(n);
              int cnt=0;
              while(!check(a)){
                  for(int i=cnt%2;i<n-1;i+=2){
                      if(a[i]>a[i+1])
                          swap(a,i,i+1);
                  }
                  cnt++;
              }
              pw.println(cnt);
          }
      }
      static void swap(int a[],int i,int j){
          int temp=a[i];
          a[i]=a[j];
          a[j]=temp;
      }
      static boolean check(int a[]){
          int n=a.length;
          for(int i=1;i<n;i++){
              if(a[i]<a[i-1])
                  return false;
          }
          return true;
      }
      static void assignAnc(ArrayList<Integer> ar[],int sz[], int pa[] ,int curr, int par){
          sz[curr] = 1;
          pa[curr] = par;
          for(int v:ar[curr]){
              if(par==v)
                  continue;
              assignAnc(ar, sz, pa, v, curr);
              sz[curr] += sz[v];
          }
      }
  
      static int findLCA(int a, int b, int par[][], int depth[]){
          if(depth[a]>depth[b]){
              a = a^b;
              b = a^b;
              a = a^b;
          }
          int diff = depth[b] - depth[a];
          for(int i=19;i>=0;i--){
              if((diff&(1<<i))>0){
                  b = par[b][i];
              }
          }
          if(a==b)
              return a;
          for(int i=19;i>=0;i--){
              if(par[b][i]!=par[a][i]){
                  b = par[b][i];
                  a = par[a][i];
              }
          }
          return par[a][0];
      }
  
      static void formArrayForBinaryLifting(int n, int par[][]){
          for(int j=1;j<20;j++){
              for(int i=0;i<n;i++){
                  if(par[i][j-1]==-1)
                      continue;
                  par[i][j] = par[par[i][j-1]][j-1];
              }
          }
      }
  
      static long lcm(int a, int b){
          return a*b/gcd(a,b);
      }
  
      static class Pair1 {
          long a;
          long b;
          
          Pair1(long a, long b) {
              this.a = a;
              this.b = b;
          }
      }
  
      static class Pair implements Comparable<Pair> {
          int a;
          int b;
          // int c;
          
          Pair(int a, int b) {
              this.a = a;
              this.b = b;
              // this.c = c;
          }
   
          public int compareTo(Pair p) {
              // if(a!=p.a)
              //     return a-p.a;
              return b-p.b;
              // return p.c - c;
          }
      }
  
      static boolean isPrime(int n) { 
          if (n <= 1) 
              return false; 
          if (n <= 3) 
              return true; 
          if (n % 2 == 0 ||  n % 3 == 0) 
              return false; 
          for (int i = 5; i * i <= n; i = i + 6) 
              if (n % i == 0 || n % (i + 2) == 0) 
                  return false; 
          return true; 
      } 
  
      static long gcd(long a, long b) {
          if (b == 0)
              return a;
          return gcd(b, a % b);
      }
  
      static long fast_pow(long base, long n, long M) {
          if (n == 0)
              return 1;
          if (n == 1)
              return base % M;
          long halfn = fast_pow(base, n / 2, M);
          if (n % 2 == 0)
              return (halfn * halfn) % M;
          else
              return (((halfn * halfn) % M) * base) % M;
      }
  
      static long modInverse(long n, long M) {
          return fast_pow(n, M - 2, M);
      }
  
      static class InputReader {
          public BufferedReader reader;
          public StringTokenizer tokenizer;
  
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 9992768);
              tokenizer = null;
          }
  
          public String next() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
  
          public int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public long nextLong() {
              return Long.parseLong(next());
          }
      }
  }