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++ (GCC 9.2.1) (judge verdict: Accepted)

//#define _GLIBCXX_DEBUG
#include<bits/stdc++.h>
using namespace std;
template <class A, class B> inline bool chmax(A &a, const B &b) { return b > a && (a = b, true); }
template <class A, class B> inline bool chmin(A &a, const B &b) { return b < a && (a = b, true); }
typedef long long ll;
typedef vector<int> vint;
typedef pair<int, int> pint;
typedef vector<long long> vlong;
typedef vector<string> vstring;
#define vpush(a,x) a.push_back(x);
#define rep(i, n) REP(i, 0, n)
#define all(v) v.begin(), v.end()
#define REP(i, x, n) for(int i = x; i < n; i++)
const int INF = 1 << 30;
const int dx[] = {1,0,-1,0,1,1,-1,-1};
const int dy[] = {0,-1,0,1,1,-1,-1,1};
#define stp(x)  setprecision(x)
const long double PI=3.14159265358979323846264338327950;
int main(){
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n;
  cin>>n;
  vint a(n);
  set<pair<int,int> > S;
  rep(i,n){
    cin>>a[i];
    S.insert(make_pair(a[i],(i+1)%2));
  }
  int count=0;
  sort(all(a));
  rep(i,n){
    if(S.find(make_pair(a[i],(i+1)%2))==S.end()){
      count++;
    }
    i++;
  }
  cout<<count<<'\n';
  return(0);
}