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>
const long long INF = 1LL << 60;
const long long MOD = 1000000007;
const double PI = acos(-1.0);
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rep1(i, n) for (ll i = 1; i <= (n); ++i)
#define rrep(i, n) for (ll i = (n - 1); i >= 0; --i)
#define perm(c) sort(ALL(c));for(bool c##p=1;c##p;c##p=next_permutation(ALL(c)))
#define ALL(obj) (obj).begin(), (obj).end()
#define RALL(obj) (obj).rbegin(), (obj).rend()
#define pb push_back
#define to_s to_string
#define len(v) (ll)v.size()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define print(x) cout << (x) << '\n'
#define drop(x) cout << (x) << '\n', exit(0)
#define debug(x) cout << #x << ": " << (x) << '\n'
using namespace std;
using ll = long long;
typedef pair<ll, ll> P;
typedef vector<ll> vec;
typedef vector<vector<ll>> vec2;
typedef vector<vector<vector<ll>>> vec3;
template<class S, class T> inline bool chmax(S &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class S, class T> inline bool chmin(S &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline ll msb(ll v) { return 1 << (31 - __builtin_clzll(v)); }
inline ll devc(ll x, ll y) { return (x + y - 1) / y; }
inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }

struct IoSetup {
    IoSetup() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(10);
        cerr << fixed << setprecision(10);
    }
} iosetup;

template< typename T1, typename T2 >
ostream &operator << (ostream &os, const pair< T1, T2 > &p) {
    os << p.first << " " << p.second;
    return os;
}

template< typename T1, typename T2 >
istream &operator >> (istream &is, pair< T1, T2 > &p) {
    is >> p.first >> p.second;
    return is;
}

template< typename T1, typename T2, typename T3 >
ostream &operator << (ostream &os, const tuple< T1, T2, T3 > &t) {
    os << get<0>(t) << " " << get<1>(t) << " " << get<2>(t);
    return os;
}

template< typename T1, typename T2, typename T3 >
istream &operator >> (istream &is, tuple< T1, T2, T3 > &t) {
    is >> get<0>(t) >> get<1>(t) >> get<2>(t);
    return is;
}

template< typename T >
ostream &operator << (ostream &os, const vector< T > &v){
    for (int i = 0; i < (int)v.size(); ++i) {
        os << v[i] << (i + 1 != v.size() ? " " : "");
    }
    return os;
}

template< typename T >
istream &operator >>  (istream &is, vector< T > &v){
    for(T &in : v) is >> in;
    return is;
}

/*---------------------------------    Tools    ------------------------------------------*/
template< typename T >
vector<T> cumsum(const vector<T> &X){
    vector<T> res(X.size() + 1, 0);
    for(int i = 0; i < X.size(); ++i) res[i + 1] += res[i] + X[i];
    return res;
}

template< typename S, typename T, typename F>
pair<T, T> bisearch(S left, T right, F f) {
    while(abs(right - left) > 1){
        T mid = (right + left) / 2;
        if(f(mid)) right = mid;
        else left = mid;
    }
    return {left, right};
}

template< typename S, typename T, typename F>
double trisearch(S left, T right, F f, int maxLoop = 90){
    double low = left, high = right;
    while(maxLoop--){
        double mid_left = high / 3 + low * 2 / 3;
        double mid_right = high * 2 / 3 + low / 3;
        if(f(mid_left) >= f(mid_right)) low = mid_left;
        else high = mid_right;
    }
    return (low + high) * 0.5;
}

/*------------------------------- Main Code Here -----------------------------------------*/

int main()
{
    ll N;
    cin >> N;
    vec A(N);
    cin >> A;

    priority_queue<ll, vec, greater<>> pqf, pqs;
    rep(i, N){
        if(i % 2 == 0) pqf.push(A[i]);
        else pqs.push(A[i]);
    }

    ll ct = 0;
    while(!pqf.empty() and !pqs.empty()){
        ll f = pqf.top(), s = pqs.top();
        pqf.pop(), pqs.pop();
        if(f < s) continue;

        pqf.push(s), pqs.push(f);
        ++ct;
    }
    print(ct);

    return 0;
}