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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
#define MOD ((int)(1e9) + 7)

const int N = ((int)1e5) + 5;

int n;
int arr[N];
int narr[N];
int ans;
map<int,int> ind;

int main ()
{
	scanf("%d",&n);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d",&arr[i]);
		narr[i] = arr[i];
		ind[arr[i]] = i % 2;
	}
	sort(narr + 1, narr + n + 1);
/*	for (int w = 1; w <= 100; w++)
	{
		for (int i = 1; i <= n; i++)
		{
			if (arr[i + 2] == -1)
				break;
			if (arr[i] > arr[i + 2])
				swap(arr[i],arr[i + 2]);
		}
	}*/
	for (int i = 1; i <= n; i++)
	{
		if (ind[narr[i]] != ind[arr[i]])
		{
			ans++;
		}
	}
	if (ans == 0)
		return printf("0"),0;
	printf("%d",ans / 2);
}