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 <algorithm>
#include <cstdio>
#include <ctype.h>
#include <functional>
#include <iostream>
#include <cassert>
#include <cfloat>
#include <climits>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <time.h>
#include <vector>
#define int long long
using namespace std;
long long mergecount(vector<int> &a) {
	long long count = 0;
	long long n = a.size();
	if (n > 1) {
		vector<int> b(a.begin(), a.begin() + n / 2);
		vector<int> c(a.begin() + n / 2, a.end());
		count += mergecount(b);
		count += mergecount(c);
		for (int i = 0, j = 0, k = 0; i < n; ++i)
			if (k == c.size())       a[i] = b[j++];
			else if (j == b.size())  a[i] = c[k++];
			else if (b[j] <= c[k])   a[i] = b[j++];
			else { a[i] = c[k++]; count += n / 2 - j; }
	}
	return count;
}
signed main() {
	
	int N;
	cin >> N;
	vector <int> A(N);
	vector <int> B;
	vector <int> C;

	
	for (int i = 0; i < N; i++) {
		cin >> A[i];
		if (i % 2 == 0) {
			B.push_back(A[i]);
		}
		else {
			C.push_back(A[i]);
		}
	}
	sort(B.begin(), B.end());
	sort(C.begin(), C.end());
	
	for (int i = 0; i < N; i++) {
		if (i % 2 == 0) {
			A[i] = B[i / 2];
		}
		else {
			A[i] = C[i / 2];
		}
	}
	
	
	cout << mergecount(A)  << endl;
	
	return 0;
}