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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

#define MAX(A, B) (((A) < (B)) ? (B) : (A))
#define ABS(A) MAX((A), (0-(A)))

int main() {
  int N;
  scanf("%d", &N);
  int** list = malloc (N * sizeof(int*));
  
  for (int i=0; i<N; i++) {
    *(list + i) = malloc(3 * sizeof(int));
    scanf("%ld", *(list + i));
    *(*(list + i) + 1) = i;
  }
  
  int cmp(const void* a, const void* b) {
    return (**(int**) a - **(int**) b);
  }
  
  qsort (list, N, sizeof(int*), cmp);
  int c=0;
  
  for (int i=0; i<N; i++) {
    c += ABS((i - *(*(list + i) + 1))) % 2;
  }
  
  c /= 2;
  printf("%d\n", c);
  
  return 0;
}