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.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        for(int i=0; i<n; i++){
            a[i] = sc.nextInt();
        }
        sc.close();

        if(n == 1){
            System.out.println(0);
            return;
        }else if(n == 2){
            if(a[0] > a[1]){
                System.out.println(1);
            }else{
                System.out.println(0);
            }
            return;
        }

        int[] a1, a2;
        if(n % 2 == 0){
            a1 = new int[n/2];
            a2 = new int[n/2];
        }else{
            a1 = new int[n/2+1];
            a2 = new int[n/2];
        }
        for(int i=0; i<a1.length; i++){
            a1[i] = a[i*2];
        }
        for(int i=0; i<a2.length; i++){
            a2[i] = a[i*2+1];
        }

        Arrays.sort(a1);
        Arrays.sort(a2);

        for(int i=0; i<n; i++){
            if(i % 2 == 0){
                a[i] = a1[i/2];
            }else{
                a[i] = a2[i/2];
            }
        }

        int pos1 = 0;
        int pos2 = 0;
        long ans = 0;
        for(int i=0; i<n; i++){
            if(pos1 >= a1.length || pos2 >= a2.length) break;
            if(a1[pos1] > a2[pos2]){
                ans += Math.max(0, pos2-pos1+1);
                pos2++;
            }else if(a1[pos1] < a2[pos2]){
                ans += Math.max(0, pos1-pos2);
                pos1++;
            }else{
                if(pos1 <= pos2){
                    pos1++;
                }else{
                    pos2++;
                }
            }
        }

        System.out.println(ans);
    }
}