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;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef unsigned char ubyte;

const ld EPS = 1e-9;
const ld PI = acos(-1.0);
const int INF = (int)1e9 + 5;
const ll LINF = (ll)1e18 + 5;

bool eq(double a, double b) {
	return fabs(a - b) < EPS;
}

template<class T>
T sqr(const T & x) { return x * x; }

template<class T> 
ll abs(const T & x) { return x < 0 ? -x : x; }

template<class T>
ll myround(const T & x) { return x < 0 ? x - 0.5 : x + 0.5; }

template<class T> 
bool chmin(T & x, const T & y) {
	if (y < x) {
		x = y;
		return true;
	}
	return false;
}

template<class T>
ostream & operator <<(ostream & os, const vector<T> & v) {
	os << "{";
	for (typename vector<T>::const_iterator it = v.begin(); it != v.end(); ++it) {
		os << *it;
		if ((it + 1) != v.end()) os << ", ";
	}
	os << "}";
	return os;
}

template<class P, class Q>
ostream & operator <<(ostream & os, const pair<P, Q> & p) {
	return os << "(" << p.first << ", " << p.second << ")";
}

template<class T> 
bool chmax(T & x, const T & y) {
	if (x < y) {
		x = y;
		return true;
	}
	return false;
}

#define all(a) (a).begin(), (a).end()

int main() {
	//srand(time(0));
#ifdef LOCAL 
	freopen("input.txt", "r", stdin);
#else
#endif

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

	vector<int> a(n), k;

	for (int & x : a) {
		scanf("%d", &x);
		k.push_back(x);
	}

	sort(all(k));

	for (int & x : a) {
		x = lower_bound(all(k), x) - k.begin();
	}

	ll res = 0;
	for (int i = 0; i < n; i++) {
		res += abs(a[i] - i) % 2;
	}

	cout << res/2 << endl;

}