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 Rust (1.15.1) (judge verdict: Accepted)


use std::io;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}


fn main() {
    let n: usize = read_line().trim().parse().unwrap();
    let mut a = Vec::with_capacity(n);
    for i in 0..n {
        let v:usize = read_line().trim().parse().unwrap();
        a.push((v, i));
    }
    a.sort();
    let mut ans = 0;
    for i in 0..n {
        if a[i].1 % 2 != i % 2 {
            ans += 1;
        }
    }

    println!("{}", ans / 2);
}