Problem:
Score : 600 points

Problem Statement

Snuke got an integer sequence of length N from his mother, as a birthday present. The i-th (1 ≦ i ≦ N) element of the sequence is a_i. The elements are pairwise distinct.
He is sorting this sequence in increasing order.
With supernatural power, he can perform the following two operations on the sequence in any order:

Operation 1: choose 2 consecutive elements, then reverse the order of those elements.

Operation 2: choose 3 consecutive elements, then reverse the order of those elements.

Snuke likes Operation 2, but not Operation 1. Find the minimum number of Operation 1 that he has to perform in order to sort the sequence in increasing order.

Constraints

1 ≦ N ≦ 10^5

0 ≦ A_i ≦ 10^9

If i ≠ j, then A_i ≠ A_j.

All input values are integers.

Input

The input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the minimum number of times Operation 1 that Snuke has to perform.

Sample Input 1

4
2
4
3
1

Sample Output 1

1

The given sequence can be sorted as follows:

First, reverse the order of the last three elements. The sequence is now: 2,1,3,4.

Then, reverse the order of the first two elements. The sequence is now: 1,2,3,4.

In this sequence of operations, Operation 1 is performed once. It is not possible to sort the sequence with less number of Operation 1, thus the answer is 1.

Sample Input 2

5
10
8
5
3
2

Sample Output 2

0

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import static java.lang.Math.*;
import static java.util.Arrays.*;

import java.util.*;
import java.io.*;
import java.math.BigInteger;

import static java.math.BigInteger.*;

public class Main {
	void solve() {
		int N = sc.nextInt();
		int[] A = sc.nextIntArray(N);
		A = normalize(A);
		int x = 0;
		for (int i = 0; i < N; i += 2) 
			if (A[i] % 2 == 1) x++;
		out.println(x);
	}
	public static int[] normalize(int[] v) {
		int[] res = new int[v.length];
		int[][] t = new int[v.length][2];
		for (int i = 0; i < v.length; i++) {
			t[i][0] = v[i];
			t[i][1] = i;
		}
		Arrays.sort(t, 0, t.length, new Comparator<int[]>(){
			public int compare(int[] a, int[] b){
				if (a[0] != b[0]) return a[0] < b[0] ? -1 : 1;
				return 0;
			}
		});

		int r = 0;
		for (int i = 0; i < v.length; i++) {
			r += (i > 0 && t[i - 1][0] != t[i][0]) ? 1 : 0;
			res[(int)t[i][1]] = r;
		}
		return res;
	}
	
	static void tr(Object... os) { System.err.println(deepToString(os)); }

	public static void main(String[] args) throws Exception {
		new Main().run();
	}

	MyScanner sc = null;
	PrintWriter out = null;
	public void run() throws Exception {
		sc = new MyScanner(System.in);
		out = new PrintWriter(System.out);
		for (;sc.hasNext();) {
			solve();
			out.flush();
		}
		out.close();
	}

	class MyScanner {
		String line;
		BufferedReader reader;
		StringTokenizer tokenizer;

		public MyScanner(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}
		public void eat() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					line = reader.readLine();
					if (line == null) {
						tokenizer = null;
						return;
					}
					tokenizer = new StringTokenizer(line);
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
		}
		public String next() {
			eat();
			return tokenizer.nextToken();
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		public boolean hasNext() {
			eat();
			return (tokenizer != null && tokenizer.hasMoreElements());
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
	}
}