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>
#include <assert.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define PB push_back
#define MP make_pair
#define MOD 1000000007LL
#define endl "\n"
const ll UNDEF = -1;
const ll INF=1e18;
template<typename T> inline bool chkmax(T &aa, T bb) { return aa < bb ? aa = bb, true : false; }
template<typename T> inline bool chkmin(T &aa, T bb) { return aa > bb ? aa = bb, true : false; }
const ll mn=1e5+4;
pair<ll,ll> a[mn];
int main() {
	ios_base::sync_with_stdio(false);
	ll n; scanf("%lld",&n);
	for (ll i=0;i<n;i++) {
		ll x;scanf("%lld",&x);
		a[i]=MP(x,i);
	}
	sort(a,a+n);
	ll need=0;
	for (ll i=0;i<n;i++) {
		if ((a[i].second-i)%2) {
			need++;
		}
	}
	ll ans=need/2;
	printf("%lld\n",ans);
}