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: Runtime Error)

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <map>
#include <queue>
using ll = long long;
using graph = std::vector<std::vector<ll>>;

using namespace std;

int main() {
    ll N;
    cin >> N;
    vector<ll> A(0);
    vector<ll> B(0);
    for(ll i = 0; i < N; i++){
        if(i % 2 == 0){
            ll tmp = 0;
            cin >> tmp;
            A.push_back(tmp);
        }else{
            ll tmp = 0;
            cin >> tmp;
            B.push_back(tmp);
        }
    }
    sort(A.begin(),A.end());
    sort(B.begin(),B.end());
    ll count = 0;
    for(int i = 0; i < N ; i++){
        if(i % 2 == 0){
            if(A[i / 2] <= B[i / 2]){
                continue;
            }
            count++;
        }else{
            if(B[(i - 1) / 2] <= A[(i + 1) / 2]){
                continue;
            }
            count++;
        }
    }
    cout << count << endl;
}