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: Wrong Answer)

#include <iostream>
#include <map>
#include <cstdio>
#include <cstring>
#include <cmath>

//#define DEBUG_PRINT

using namespace std;

typedef unsigned long long ull;
typedef signed long long sll;

map<int, int> mp;

int main()
{
    int N;
    cin >> N;
    
    for (int i = 0; i < N; i++)
    {
        int temp;
        cin >> temp;
        mp[temp]++;
    }

#ifdef DEBUG_PRINT
    printf("-----\n");
    for (auto itr = mp.begin(); itr != mp.end(); ++itr)
    {
        printf("key=%d, value=%d\n", itr->first, itr->second);
    }
#endif
    
    bool isDifferent = false;
    while (!isDifferent)
    {
        isDifferent = true;

#ifdef DEBUG_PRINT
        printf("-----\n");
#endif
        for (auto itr = mp.begin(); itr != mp.end(); ++itr)
        {
#ifdef DEBUG_PRINT
            printf("key=%d, value=%d\n", itr->first, itr->second);
#endif
            if (itr->second >= 2)
            {
                isDifferent = false;
                itr->second--;
                
                bool isFound = false;
                for (auto pair = mp.begin(); pair != mp.end(); ++pair)
                {
                    if (pair == itr)
                    {
                        continue;
                    }
                    else if (pair->second >= 2)
                    {
                        pair->second--;
                        isFound = true;
                        break;
                    }
                }
                
                // 2 以上の要素が見つからなかったら次の1以上ある要素から -1
                if (!isFound)
                {
                    auto victim = itr;
                    for (; victim != mp.end(); ++victim)
                    {
                        if (victim->second == 1)
                        {
                            mp.erase(victim);
                        }
                    }
                }
                break;
            }
        }
    }
    
    int ans = 0;
    for (auto itr = mp.begin(); itr != mp.end(); ++itr)
    {
        ans += itr->second;
    }
    
    printf("%d\n", ans);
    
    return 0;
}