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 <string.h>
#include <stdio.h>
#include <map>
#include <vector>
#include <math.h>
#include <algorithm>
#include <queue>
#include <set>
#include <tuple>
using namespace std;

#define FOR(i,init,a) for(int i=init; i<a; i++)
#define rep(i,a) FOR(i,0,a)
#define rrep(i,a) for(int i=a; i>=0; i--)
#define rep1(i,a) for(int i=1; i<=a; i++)
#define cout1(a) cout << a << endl;
#define cout2(a,b) cout << a << " " << b << endl;
#define cout3(a,b,c) cout << a << " " << b << " " << c << endl;
#define cout4(a,b,c,d) cout << a << " " << b << " " << c << " " << d << endl;
#define mem(a,n) memset( a, n, sizeof(a))
#define all(a) a.begin(),a.end()

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> V;
typedef vector<V> VV;
typedef vector<VV> VVV;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LLINF = 1e18;
static const double pi = 3.141592653589793;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int N;
    cin>>N;
    
    vector<int> nums(N), sorted(N);
    rep(i,N) cin>>nums[i], sorted[i]=nums[i];
    
    sort(all(sorted));
    map<int,int> dic;
    rep(i,N){
        dic[sorted[i]]=(i%2);
    }
    
    int cnt=0;
    rep(i,N){
        if((i%2)!=dic[nums[i]]) cnt++;
    }
    
    cout1(cnt/2);
}