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: Wrong Answer)

#include <bits/stdc++.h>
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

#define forn(i,n) for (int i = 0; i < int(n); ++i)
#define forsn(i,x,y) for(long long i = (x) ; i <= (y) ; ++i)
typedef long long ll;
typedef long double ld;

int main() {
  vector<int> a,b;
  int n;
  cin >> n;
  forn(i,n) {
    int t;
    cin >> t;
    if(i%2 == 0) { 
      a.push_back(t);
      //a.push_back(i+5000000);
    }
    else {
      b.push_back(t);
      //b.push_back(i);
    }
  }
  sort(a.begin(),a.end());
  sort(b.begin(),b.end());
  
  int ai = 0;
  int bi = 0;
  ll total = 0;
  forn(i,n) {
    //cout << i << " " << a[ai] << " " << b[bi] << endl;
    int bp = bi * 2 + 1;
    if(bi >= b.size() || (ai < a.size() && a[ai] < b[bi]) ) {
      ai++;
    }
    else { // bi < b.size() && b[bi] < a[ai] || ai >= a.size()
      //cout << "  " << abs(i-bp) << endl;
      total += abs(i - bp);
      bi++;
    }
  }
  cout << total << endl;
  return 0;
}