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 ms_valtype int

//比較関数
int compare_ms(ms_valtype a, ms_valtype b){
	return a - b;
}

//昇順
void sort_sub(ms_valtype *origin, int left, int right, ms_valtype *tmp){
	if(right - left > 1){
		int i, j, k, half = (left + right) / 2;
		sort_sub(origin, left, half, tmp);
		sort_sub(origin, half, right, tmp);
		for(i = left; i < right; i++){
			tmp[i] = origin[i];
		}
		for(i = left, j = left, k = half; i < right; i++){
			if(k == right){
				origin[i] = tmp[j];
				j++;
			}
			else if(compare_ms(tmp[j], tmp[k]) <= 0 && j < half){
				origin[i] = tmp[j];
				j++;
			}
			else{
				origin[i] = tmp[k];
				k++;
			}
		}
	}
}

void sort(ms_valtype *origin, int N){
	ms_valtype *tmp = (ms_valtype *)malloc(sizeof(ms_valtype) * N);
	sort_sub(origin, 0, N, tmp);
	free(tmp);
}

int main(){
	int N, i, ans = 0;
	scanf("%d", &N);
	int *A = (int *)malloc(sizeof(int) * N);
	int *B = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%d", &A[i]);
		B[i] = A[i];
	}
	sort(B, N);
	int l, h, r;
	for(i = 0; i < N; i += 2){
		l = -1;
		r = N;
		while(r - l > 1){
			h = (l + r) / 2;
			if(B[h] >= A[i]){
				r = h;
			}
			else{
				l = h;
			}
		}
		ans += r % 2;
	}
	printf("%d\n", ans);
	return 0;
}