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)

//Khodaya aghideye man ra az dast oghde am masoon dar!
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
template <typename T> using ordered_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T> using ordered_multiset =  tree<T, null_type, greater_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
#pragma optimize O3
#define debug(x) cerr << #x << " : " << x;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1e5 + 7;
const int infint = (ll)1e9;
const ll inf = (ll)1e18;
ll a[MAXN], n, b[MAXN];
unordered_map<ll, ll> M;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int n;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		b[i] = a[i];
	}
	sort(b, b + n);
	for (int i = 0; i < n; i++)
		M[b[i]] = i;
	ll ans = 0;
	for (int i = 0; i < n; i++)
		if(i % 2 != M[a[i]] % 2)
			ans++;
	cout << ans / 2;
}