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>
using namespace std;
#define sz(x) ((int) (x).size())
#define forn(i,n) for (int i = 0; i < int(n); ++i)
#define all(x) (x).begin(), (x).end()
typedef long long ll;
typedef long long i64;
typedef long double ld;
const int inf = int(1e9) + int(1e5);
const ll infl = ll(2e18) + ll(1e10);

const int maxn = 100100;
int a[maxn];
int b[maxn];
int c[maxn];

int f[maxn];
int get(int r) {
    int res = 0;
    for (int i = r - 1; i >= 0; i = (i & (i + 1)) - 1)
        res += f[i];
    return res;
}
void put(int r) {
    for (int i = r; i < maxn; i |= i + 1)
        f[i]++;
}

int main() {
    #ifdef LOCAL
    assert(freopen("c.in", "r", stdin));
    #else
    #endif
    int n;
    cin >> n;
    forn (i, n) {
        if (i % 2)
            cin >> b[i / 2];
        else
            cin >> a[i / 2];
    }
    sort(a, a + (n + 1) / 2);
    sort(b, b + n / 2);
    vector<int> val;
    forn (i, n) {
        if (i % 2)
            c[i] = b[i / 2];
        else
            c[i] = a[i / 2];
        val.push_back(c[i]);
    }
    sort(all(val));
    ll res = 0;
    for (int i = n - 1; i >= 0; --i) {
        int x = lower_bound(all(val), c[i]) - val.begin();
        res += get(x);
        put(x);
    }
    cout << res << '\n';
}