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: Wrong Answer)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Collections;
import java.util.NoSuchElementException;

public class Main {
	
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		Integer[] e = new Integer[n / 2 + n % 2];
		Integer[] o = new Integer[n / 2];
		
		for (int i = 0; i < n; i++) {
			if (i % 2 == 0) {
				e[i / 2] = sc.nextInt();
			} else {
				o[i / 2] = sc.nextInt();
			}
		}
		
		Arrays.sort(e, Collections.reverseOrder());
		Arrays.sort(o, Collections.reverseOrder());
		
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			if (i % 2 == 0) {
				a[i] = e[i / 2];
			} else {
				a[i] = o[i / 2];
			}
		}
		
		int count = 0;
		for (int i = 0; i < n - 1; i++) {
			for (int j = 0; j < n - 1 - i; j++) {
				if (a[j] < a[j + 1]) {
					int tmp = a[j + 1];
					a[j + 1] = a[j];
					a[j] = tmp;
					count++;
				}
			}
		}
		
		out.println(count);
		out.flush();
	}
}

class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;
	
	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}
	
	private byte readByte() {
		if (hasNextByte()) {
			return buffer[ptr++];
		} else {
			return -1;
		}
	}
	
	private boolean isPrintableChar(int c) {
		return '!' <= c && c <= '~';
	}
	
	private void skipUnprintable() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr])) {
			ptr++;
		}
	}
	
	public boolean hasNext() {
		skipUnprintable();
		return hasNextByte();
	}
	
	public int nextInt() {
		if (!hasNext()) {
			throw new NoSuchElementException();
		}
		int n = 0;
		boolean minus = false;
		byte b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
}