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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <cstring>
#include <numeric>
#include <algorithm>
#include <functional>
#include <array>
#include <map>
#include <queue>
#include <limits.h>
#include <set>
#include <stack>
#define REP(i,n) for(int i = 0; n > i; i++)
#define MOD 1000000007
#define accm(i) ( i%2 ? od[((i)-1)/2] : ev[(i)/2])
#define acc(i) ( i%2 ? cpod[(i-1)/2] : cpev[i/2])
using namespace std;
typedef vector<int> Ivec;
typedef pair<int, int> pii;

int main() {
	int n;
	scanf("%d", &n);

	array<int, 50100> ev,od;

	REP(i, n) {
		scanf("%d", &accm(i));
	}

	int cou = 0;
	bool ch = 1;

	while (ch) {
		ch = 0;
		/*array<int, 50100> cpev(ev), cpod(od);
		sort(cpod.begin(), cpod.begin() + n / 2);
		sort(cpev.begin(), cpev.begin() + (n + 1) / 2);
		od = cpod;
		ev = cpev;*/
		sort(od.begin(), od.begin() + n / 2);
		sort(ev.begin(), ev.begin() + (n + 1) / 2);
		int bef = ev[0];
		REP(i, n - 1) {
			int temp = bef;
			bef = accm(i + 1);
			if (temp > accm(i + 1)) {
				accm(i) = accm(i + 1);
				accm(i + 1) = temp;
				ch = 1;
				cou++;
			}
		}
	}
	printf("%d\n", cou);
	return 0;
}