Problem:
Score : 400 points

Problem Statement

Snuke has decided to play a game using cards.
He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.

He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.

Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.

Constraints

3 ≦ N ≦ 10^{5}

N is odd.

1 ≦ A_i ≦ 10^{5}

A_i is an integer.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_{N}

Output

Print the answer.

Sample Input 1

5
1 2 1 3 7

Sample Output 1

3

One optimal solution is to perform the operation once, taking out two cards with 1 and one card with 2. One card with 1 and another with 2 will be eaten, and the remaining card with 1 will be returned to deck. Then, the values written on the remaining cards in the deck will be pairwise distinct: 1, 3 and 7.

Sample Input 2

15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1

Sample Output 2

7

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
 
using namespace std;
 
#define INF 1001000100010001000
#define MOD 1000000007
#define EPS 1e-10
#define int long long
#define rep(i, N) for (int i = 0; i < N; i++)
#define Rep(i, N) for (int i = 1; i < N; i++)
#define For(i, a, b) for (int i = (a); i < (b); i++)
#define pb push_back
#define mp make_pair
#define i_i pair<int, int>
#define vi vector<int>
#define vvi vector<vi >
#define vb vector<bool>
#define vvb vector<vb >
#define vp vector< i_i >
#define all(a) (a).begin(), (a).end()
#define Int(x) int x; scanf("%lld", &x);
 
//int dxy[5] = {0, 1, 0, -1, 0};
// assign
 
signed main()
{
    Int(n);

    vi c(n);

    rep(i, n) {
        cin >> c[i];
    }

    sort(all(c));

    int pt = 1, bef = c[0], cnt = 1;

    vp cs;

    while (pt < n) {

        if (bef != c[pt]) {
            cs.pb(mp(bef, cnt));
            cnt = 1;
            bef = c[pt++];
        } else {
            cnt++;
            pt++;
        }

    }

    cs.pb(mp(bef, cnt));

    int ret = 0, less = 0;

    rep(i, cs.size()) {
        if (cs[i].second % 2) {
            ret++;
        } else {
            less++;
        }
    }

    cout << ret + less - (less % 2) << endl;

    return 0;
}