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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int mergecount(vector<long> &a) {
  long count = 0;
  int n = a.size();
  if (n > 1) {
    vector<long> b(a.begin(), a.begin() + n/2);
    vector<long> c(a.begin() + n/2, a.end());
    count += mergecount(b);
    count += mergecount(c);
    for (int i = 0, j = 0, k = 0; i < n; ++i)
      if (k == c.size())       a[i] = b[j++];
      else if (j == b.size())  a[i] = c[k++];
      else if (b[j] <= c[k])   a[i] = b[j++];
      else                   { a[i] = c[k++]; count += n/2 - j; }
  }
  return count;
}

int main(int argc, char const *argv[])
{
  int N;
  cin >> N;
  vector<long> A(N-N/2);
  vector<long> B(N/2);
  vector<long> C(N);
  for (int i = 0; i < N/2; ++i)
  {
    cin >> A[i];
    cin >> B[i];
  }
  if (N % 2 == 1)
  {
    cin >> A[N-N/2-1];
  }

  // for (int i = 0; i < A.size(); ++i)
  // {
  //   cout << A[i] << endl;
  // }
  // cout << "A" << endl;
  // for (int i = 0; i < B.size(); ++i)
  // {
  //   cout << B[i] << endl;
  // }
  // cout << "B" << endl;

  sort(A.begin(), A.end());
  sort(B.begin(), B.end());

  for (int i = 0; i < N/2; ++i)
  {
    C[2 * i] = A[i];
    C[2 * i + 1] = B[i];
  }
  if (N % 2 == 1)
  {
    C[N-1] = A[N-N/2-1];
  }

  // for (int i = 0; i < N; ++i)
  // {
  //   cout << C[i] << endl;
  // }
  // cout << "C" << endl;

  cout << mergecount(C) << endl;
  return 0;
}