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 (Clang 3.8.0) (judge verdict: Accepted)

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

int descending_compare(const void *a, const void *b){
    if (*(int*)a > *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

int ascending_compare(const void *a, const void *b){
    if (*(int*)a < *(int*)b){
        return -1;
    }else if (*(int*)a == *(int*)b){
        return 0;
    }else{
        return 1;
    }
}

// array pointer = *a, num of element = n, key key
int lower_bound(int *a, int n, int key){
    int ng, mid, ok;
    ng = -1, ok = n-1;
    while (abs(ok - ng) > 1){
        mid = (ok + ng) / 2;
        if (key <= a[mid]){
            ok = mid;
        }else{
            ng = mid;
        }
    }
    if (a[ok] >= key)return ok;
    return n;
}

//greatest common divisor
unsigned long gcd(unsigned long x, unsigned long y){
    if (y == 0){ 
        return x;
    }else if (x > y){
        return gcd(y, x % y);
    }else{
        return gcd(x, y % x);
    }
}

unsigned long lcm(unsigned long x, unsigned long y){
    unsigned long g = gcd(x, y);
    return x*y/g;

}



long long factorial(int x){
    long long rtn = 1;
    int i;
    for (i = x; i > 1; i--){
        rtn = (rtn*i);
    }
    return rtn;
}

/*
int struct_ascending_compare(const void *p, const void *q) {
return ((struct_name*)p)->member_name - ((struct_name*)q)->member_name;
}*/

/*unsigned long long pascal[100][100] = {0};
void make_pascal(void){
    for (int i = 0; i < 100; i++){
        pascal[i][0] = 1;
    }
    pascal[1][1] = 1;
    for (int i = 2; i < 100; i++){
        for (int j = 1; j < 100; j++){
           pascal[i][j] = (pascal[i-1][j-1]+pascal[i-1][j]) % mod;
        }
    }
}*/
long long mod = 1000000007;

//x ^ n
long long mod_pow(long long x, long long n){
    long long res = 1;
    for(int i = 0;i < 60; i++){
        if(n >> i & 1) res = res * x % mod;
        x = x * x % mod;
    }
    return res;
}

int hindo[100005];

int main(void){
    int n;
    int a;
    int to_eat = 0;
    int to_remain = 0;
    scanf("%d", &n);

    for (int i = 0; i < n; i++){
        scanf("%d", &a);
        if (hindo[a]) to_eat++;
        else{
            hindo[a] = 1;
            to_remain++;
        }
    }
    if (to_eat % 2){
        printf("%d\n", to_remain - 1);
    }else{
        printf("%d\n", to_remain);
    }
    return 0;

}