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.util.*;
  import java.io.*;
  
  public class A {
  	static StringBuilder sb;
  	static dsu dsu;
  	static long fact[];
  	static int mod = (int) (1e9 + 7);
  
  	static void solve() {
  		int n = i();
  		int[] arr = readArrayi(n);
  
  		int ans = 0;
  		for (int i = 0; i < n; i++) {
  			if (i % 2 == 0) {
  				for (int j = 0; j < n - 2; j += 2) {
  					if (arr[j] > arr[j + 1]) {
  						int temp = arr[j];
  						arr[j] = arr[j + 1];
  						arr[j + 1] = temp;
  						ans = i + 1;
  					}
  				}
  			} else {
  				for (int j = 1; j < n - 1; j += 2) {
  					if (arr[j] > arr[j + 1]) {
  						int temp = arr[j];
  						arr[j] = arr[j + 1];
  						arr[j + 1] = temp;
  						ans = i + 1;
  					}
  				}
  			}
  		}
  
  		sb.append(ans + "\n");
  	}
  
  	public static void main(String[] args) {
  		sb = new StringBuilder();
  		int test = 1;
  		test = i();
  		while (test-- > 0) {
  			solve();
  		}
  		System.out.println(sb);
  
  	}
  
  	/*
  	 * fact=new long[(int)1e6+10]; fact[0]=fact[1]=1; for(int i=2;i<fact.length;i++)
  	 * { fact[i]=((long)(i%mod)1L(long)(fact[i-1]%mod))%mod; }
  	 */
  //**************NCR%P******************	 
  	static long ncr(int n, int r) {
  		if (r > n)
  			return (long) 0;
  
  		long res = fact[n] % mod;
  		// System.out.println(res);
  		res = ((long) (res % mod) * (long) (p(fact[r], mod - 2) % mod)) % mod;
  		res = ((long) (res % mod) * (long) (p(fact[n - r], mod - 2) % mod)) % mod;
  		// System.out.println(res);
  		return res;
  
  	}
  
  	static long p(long x, long y)// POWER FXN //
  	{
  		if (y == 0)
  			return 1;
  
  		long res = 1;
  		while (y > 0) {
  			if (y % 2 == 1) {
  				res = (res * x) % mod;
  				y--;
  			}
  
  			x = (x * x) % mod;
  			y = y / 2;
  
  		}
  		return res;
  	}
  
  //**************END******************
  
  	// *************Disjoint set
  	// union*********//
  	static class dsu {
  		int parent[];
  
  		dsu(int n) {
  			parent = new int[n];
  			for (int i = 0; i < n; i++)
  				parent[i] = -1;
  		}
  
  		int find(int a) {
  			if (parent[a] < 0)
  				return a;
  			else {
  				int x = find(parent[a]);
  				parent[a] = x;
  				return x;
  			}
  		}
  
  		void merge(int a, int b) {
  			a = find(a);
  			b = find(b);
  			if (a == b)
  				return;
  			parent[b] = a;
  		}
  	}
  
  //**************PRIME FACTORIZE **********************************//
  	static TreeMap<Integer, Integer> prime(long n) {
  		TreeMap<Integer, Integer> h = new TreeMap<>();
  		long num = n;
  		for (int i = 2; i <= Math.sqrt(num); i++) {
  			if (n % i == 0) {
  				int nt = 0;
  				while (n % i == 0) {
  					n = n / i;
  					nt++;
  				}
  				h.put(i, nt);
  			}
  		}
  		if (n != 1)
  			h.put((int) n, 1);
  		return h;
  
  	}
  
  //****CLASS PAIR ************************************************
  	static class Pair implements Comparable<Pair> {
  		int x;
  		int y;
  
  		Pair(int x, int y) {
  			this.x = x;
  			this.y = y;
  		}
  
  		public int compareTo(Pair o) {
  			return (int) (this.y - o.y);
  
  		}
  
  	}
  //****CLASS PAIR **************************************************
  
  	static class InputReader {
  		private InputStream stream;
  		private byte[] buf = new byte[1024];
  		private int curChar;
  		private int numChars;
  		private SpaceCharFilter filter;
  
  		public InputReader(InputStream stream) {
  			this.stream = stream;
  		}
  
  		public int read() {
  			if (numChars == -1)
  				throw new InputMismatchException();
  			if (curChar >= numChars) {
  				curChar = 0;
  				try {
  					numChars = stream.read(buf);
  				} catch (IOException e) {
  					throw new InputMismatchException();
  				}
  				if (numChars <= 0)
  					return -1;
  			}
  			return buf[curChar++];
  		}
  
  		public int Int() {
  			int c = read();
  			while (isSpaceChar(c))
  				c = read();
  			int sgn = 1;
  			if (c == '-') {
  				sgn = -1;
  				c = read();
  			}
  			int res = 0;
  			do {
  				if (c < '0' || c > '9')
  					throw new InputMismatchException();
  				res *= 10;
  				res += c - '0';
  				c = read();
  			} while (!isSpaceChar(c));
  			return res * sgn;
  		}
  
  		public String String() {
  			int c = read();
  			while (isSpaceChar(c))
  				c = read();
  			StringBuilder res = new StringBuilder();
  			do {
  				res.appendCodePoint(c);
  				c = read();
  			} while (!isSpaceChar(c));
  			return res.toString();
  		}
  
  		public boolean isSpaceChar(int c) {
  			if (filter != null)
  				return filter.isSpaceChar(c);
  			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
  		}
  
  		public String next() {
  			return String();
  		}
  
  		public interface SpaceCharFilter {
  			public boolean isSpaceChar(int ch);
  		}
  	}
  
  	static class OutputWriter {
  		private final PrintWriter writer;
  
  		public OutputWriter(OutputStream outputStream) {
  			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
  		}
  
  		public OutputWriter(Writer writer) {
  			this.writer = new PrintWriter(writer);
  		}
  
  		public void print(Object... objects) {
  			for (int i = 0; i < objects.length; i++) {
  				if (i != 0)
  					writer.print(' ');
  				writer.print(objects[i]);
  			}
  		}
  
  		public void printLine(Object... objects) {
  			print(objects);
  			writer.println();
  		}
  
  		public void close() {
  			writer.close();
  		}
  
  		public void flush() {
  			writer.flush();
  		}
  	}
  
  	static InputReader in = new InputReader(System.in);
  	static OutputWriter out = new OutputWriter(System.out);
  
  	public static long[] sort(long[] a2) {
  		int n = a2.length;
  		ArrayList<Long> l = new ArrayList<>();
  		for (long i : a2)
  			l.add(i);
  		Collections.sort(l);
  		for (int i = 0; i < l.size(); i++)
  			a2[i] = l.get(i);
  		return a2;
  	}
  
  	public static int[] sort(int[] a2) {
  		int n = a2.length;
  		ArrayList<Integer> l = new ArrayList<>();
  		for (int i : a2)
  			l.add(i);
  		Collections.sort(l);
  		for (int i = 0; i < l.size(); i++)
  			a2[i] = l.get(i);
  		return a2;
  	}
  
  	public static long pow(long x, long y) {
  		long res = 1;
  		while (y > 0) {
  			if (y % 2 != 0) {
  				res = (res * x);// % modulus;
  				y--;
  
  			}
  			x = (x * x);// % modulus;
  			y = y / 2;
  		}
  		return res;
  	}
  
  //GCD___+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
  	public static long gcd(long x, long y) {
  		if (x == 0)
  			return y;
  		else
  			return gcd(y % x, x);
  	}
  	// ******LOWEST COMMON MULTIPLE
  	// *********************************************
  
  	public static long lcm(long x, long y) {
  		return (x * (y / gcd(x, y)));
  	}
  
  //INPUT PATTERN********************************************************
  	public static int i() {
  		return in.Int();
  	}
  
  	public static long l() {
  		String s = in.String();
  		return Long.parseLong(s);
  	}
  
  	public static String s() {
  		return in.String();
  	}
  
  	public static int[] readArrayi(int n) {
  		int A[] = new int[n];
  		for (int i = 0; i < n; i++) {
  			A[i] = i();
  		}
  		return A;
  	}
  
  	public static long[] readArray(long n) {
  		long A[] = new long[(int) n];
  		for (int i = 0; i < n; i++) {
  			A[i] = l();
  		}
  		return A;
  	}
  
  }