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: Wrong Answer)

# include <iostream>
# include <cmath>
# include <algorithm>
# include <stdio.h>
# include <cstdint>
# include <cstring>
# include <string>
# include <cstdlib>
# include <vector>
# include <bitset>
# include <map>
# include <queue>
# include <ctime>
# include <stack>
# include <set>
# include <list>
# include <random>
# include <chrono>
# include <deque>
# include <functional>
# include <iomanip>
# include <sstream>
# include <fstream>
# include <complex>
# include <numeric>
# include <immintrin.h>
# include <cassert>
# include <array>
# include <tuple>
# include <cctype>
# include <unordered_map>
# include <unordered_set>
//#include <bits/stdc++.h>
 
#pragma GCC optimize("Ofast")
//#pragma GCC optimization("unroll-loops, no-stack-protector")
//#pragma GCC target("avx,avx2,fma")
 
#define forn(i, n) for (int i = 0; i < (n); i++)
#define forx(i,x,n) for (int i = x; i < (n); i++)
#define form(i, n) for (int i = n-1; i>=0; i--)
#define all(x) (x).begin(), (x).end()
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define pf push_front
#define mp make_pair
#define ll long long
#define ld long double
#define ull unsigned ll
#define srt(a) sort(a.begin(), a.end());
#define pll pair<ll, ll>
#define pii pair<int,int>
#define pld pair<ld,ld>
#define ar array
#define endl '\n';
 
using namespace std;
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
template <class T> inline T gcd(T a, T b) { while (b) { a %= b; swap(a, b); } return a; }
template <class T> inline T lcm(T a, T b) { return a * b / gcd(a, b); }
 
const ll MAXN = 2e5 + 7, K = 200043, M = 1000043, INF = 1e18 + 7;
// the good way to get PI number
const ld PI = acos(-1.0);
ll MOD = 1e9 + 7;

template <class T> T sqr(T a) {
    return a * a;
}

ll mul(ll a, ll b) {
    return ((a % MOD) * (b % MOD)) % MOD;
}

ll sum(ll a, ll b) {
    return ((a % MOD) + (b % MOD)) % MOD;
}

ll sub(ll a, ll b) {
    return sum(a, MOD - b);
}

ll bin_pow(ll a, ll n) {
    ll res = 1;
    while (n > 0) {
        if (n & 1)
            res = mul(res, a);
        a = mul(a, a);
        n >>= 1;
    }
    return res;
}

ll inv(ll x) {
    return bin_pow(x, MOD - 2) % MOD;
}

ll fact[MAXN];

ll C(ll n, ll k) {
    return mul(mul(fact[n], inv(fact[k])), inv(fact[n - k]));
}

void solve() {
    int n;
    cin >> n;
    
    vi a(n), even, odd;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        if (i % 2)
            odd.pb(a[i]);
        else
            even.pb(a[i]);
    }
    
    sort(even.begin(), even.end());
    sort(odd.begin(), odd.end());
    sort(a.begin(), a.end());
    
    int ans1 = 0, ans2 = 0;
    for (int i = 0; i < even.size(); ++i) {
        if (a[2*i] != even[i])
            ++ans1;
    }
    
    for (int i = 0; i < odd.size(); ++i) {
        if (a[2*i + 1] != odd[i])
            ++ans2;
    }
    
    cout << min(ans1, ans2) << endl;
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    
    int t = 1;
//    cin >> t;
    
    while (t --> 0)
        solve();
    return 0;
}