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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <stack>
#include <functional>

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; ++i)
#define repp(i, m, n) for(ll i = m, i##_len = (n); i < i##_len; ++i)
#define reprev(i, n) for(ll i = (n-1LL); i >= 0; --i)
#define all(x) (x).begin(), (x).end()
template <class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template <class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }
template <class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair <int,int> P;
typedef long double ld;

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<int> a(n), b((n + 1) / 2);
	rep (i, n) cin >> a[i];
	rep (i, (n + 1) / 2) b[i] = a[2 * i];
	sort(all(a));
	rep (i, (n + 1) / 2) eprintf("%d ", a[2 * i]); eprintf("\n");
	rep (i, (n + 1) / 2) eprintf("%d ", b[i]); eprintf("\n");
	map<int, int> m;
	rep (i, (n + 1) / 2) {
		m[a[2 * i]]++;
		m[b[i]]--;
	}
	ll ans = 0;
	for (auto x : m) {
		if (x.second > 0) ans += x.second;
	}
	cout << ans << "\n";
	return 0;
}