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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
int arr[100005];
int main()
{
	//entrada
	//freopen ("a.in" , "r" , stdin);
    int N; cin >> N;
    for (int i = 1 ; i<= N;i++) cin >> arr[i];
    int cant = 0;
    bool change = true;
    int k = 3;
while(change)
{
    //realizando todas las posibles formas con 3
    //ejecuto hasta ke no haya cambios con 3
	while(change)
    {
        change = false;
        for (int i = k -2; i <= N-2  ;i++)
        {
            if (arr[i+2]<arr[i])
            {
            int temp = arr[i];
            arr[i] = arr[i+2];
            arr[i+2] = temp;
            change = true;
            }
        }
    }
    for (int i = 1 ; i <= N-1  ;i++)
    {
        if (arr[i+1]<arr[i])
        {
            int temp = arr[i];
            arr[i] = arr[i+1];
            arr[i+1] = temp;
            cant++;
            change = true;
            if (i >= 3) k = i;
            break;
        }
    }
}
    cout << cant << endl;
    return 0;
}