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>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ll long long
#define ld long double
int gcd(int x, int y) { return (x % y)? gcd(y, x % y): y; }    //最大公約数
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }     //最小公倍数
using Graph = vector<vector<ll>>;
ll inf=300000000000000000;
const double PI = 3.14159265358979323846;
const int MAX = 510000;
const ll MOD = 1000000007;
int main(){
  int n;
  cin >> n;
  set<int> even;
  set<int> odd;
  int b[n];
  rep(i,n){
    int u;
    cin >> u;
    if(i%2==0)even.insert(u);
    else odd.insert(u);
    b[i]=u;
  }
  sort(b,b+n);
  int cnt=0;
  rep(i,n){
    int u=b[i];
    if((i%2==0&&odd.count(u))||(i%2==1&&even.count(u)))cnt++;
  }
  cout << cnt << endl;
}