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 <iostream>
#include <algorithm>
#include <vector>
using namespace std;
#define FOR(t,i,n) for(t i=(t)0;(i)<(n);(i)++)
template<class T>void printarray(T a,int n){
    FOR(int,i,n) cout << a[i] << " " ;
    cout << endl;
}
main(){
    vector<int> even,odd;
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        int x;cin>>x;
        if(i&1) even.push_back(x);
        else odd.push_back(x) ;
    } 
    sort(odd.begin(),odd.end());
    sort(even.begin(),even.end());
    int a[n],e=0,o=0;
    for(int i=0;i<n;i++){
        if(i&1) a[i] = even[e++] ;
        else a[i] = odd[o++];
    } 
    int ans=0;
  //  printarray(odd,odd.size());
  //  printarray(even,even.size()) ;
  //  printarray(a,n);
    for(int i=0;i<n-1;i++){
        if(a[i]>a[i+1]){
            ans++;
            swap(a[i],a[i+1]); 
        } 
    }
   // printarray(a,n) ;
    cout << ans << endl;
}