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)

/* template.cpp [[[ */
#include <bits/stdc++.h>
using namespace std;
#define get_macro(a, b, c, d, name, ...) name
#define rep(...) get_macro(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)
#define rrep(...) get_macro(__VA_ARGS__, rrep4, rrep3, rrep2, rrep1)(__VA_ARGS__)
#define rep1(n) rep2(i_, n)
#define rep2(i, n) rep3(i, 0, n)
#define rep3(i, a, b) rep4(i, a, b, 1)
#define rep4(i, a, b, s) for (ll i = (a); i < (ll)(b); i += (ll)s)
#define rrep1(n) rrep2(i_, n)
#define rrep2(i, n) rrep3(i, 0, n)
#define rrep3(i, a, b) rrep4(i, a, b, 1)
#define rrep4(i, a, b, s) for (ll i = (ll)(b) - 1; i >= (ll)(a); i -= (ll)s)
#define each(x, c) for (auto &&x : c)
#define fs first
#define sc second
#define all(c) begin(c), end(c)
using ui = unsigned;
using ll = long long;
using ul = unsigned long long;
using ld = long double;
const int inf = 1e9 + 10;
const ll inf_ll = 1e18 + 10;
const ll mod = 1e9 + 7;
const ll mod9 = 1e9 + 9;
const int dx[]{-1, 0, 1, 0, -1, 1, 1, -1};
const int dy[]{0, -1, 0, 1, -1, -1, 1, 1};
template<class T, class U> void chmin(T &x, const U &y){ x = min<T>(x, y); }
template<class T, class U> void chmax(T &x, const U &y){ x = max<T>(x, y); }
struct prepare_ { prepare_(){ cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(12); } } prepare__;
/* ]]] */

int n;
int a[100000];
vector<int> z, b, c;
int bit[100010];
int sum(int k){
  int s = 0;
  while (k){
    s += bit[k];
    k -= k & -k;
  }
  return s;
}
int sum(int l, int r){
  return sum(r) - sum(l);
}
void add(int k, int x){
  k++;
  while (k <= 100000){
    bit[k] += x;
    k += k & -k;
  }
}

int main(){
  cin >> n;
  rep(i, n) cin >> a[i];
  rep(i, 0, n, 2) b.push_back(a[i]);
  rep(i, 1, n, 2) c.push_back(a[i]);
  sort(all(b));
  sort(all(c));
  rep(i, b.size()) a[i * 2] = b[i];
  rep(i, c.size()) a[i * 2 + 1] = c[i];
  rep(i, n) z.push_back(a[i]);
  sort(all(z));
  z.erase(unique(all(z)), end(z));
  rep(i, n) a[i] = lower_bound(all(z), a[i]) - begin(z);
  int s = 0;
  rep(i, n){
    s += sum(a[i] + 1, 100001);
    add(a[i], 1);
  }
  cout << s << endl;
}